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A    NEW     TREATISE 

ON 

ELEMENTS  OF  MECHANICS 

ESTABLISHING    STRICT    PRECISION 

IN   THE  MEANING   OF 

DYNAMICAL    TERMS 

ACCOMPANIED   WITH    AN 

APPENDIX 

ON 

DUODENAL    ARITHMETIC    AND    METROLOGY 

BY 

JOHN    W.  (NYSTROM,   C.  E. 


NEW  YORK 

G.  P.  PUTNAM'S  SONS 

182  FIFTH  AVENUE 
1877. 


Entered  according  to  Act  of  Congress,  in  the  year  1875,  by 

JOHN  W.  NYSTROM, 
In  the  Office  of  the  Librarian  of  Congress,  at  Washington. 


filC 


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2 

276181 


TO 

WILLIAM   SELLERS,  ESQ., 

THIS  WORK  is  JR.ESPECTFULLY  DEDICATED, 

AS    A    MARK   OF    APPRECIATION    OF    HIS 

DISTINGUISHED    ACCOMPLISHMENTS 

IN  THE  MECHANIC  ARTS, 

BY  THE  AUTHOR. 


PREFACE. 


THE  principal  objects  in  introducing  this  new  treatise  on  me- 
chanics are  the  establishment  of  strict  precision  in  the  meaning 
of  dynamical  terms,  and  the  classification  of  physical  quantities 
into  elements  and  functions. 

A  revision  of  the  principles  of  mechanics  is  a  necessity  long 
felt,  and  frequently  acknowledged  in  the  discussions  of  learned 
men,  who  have  heretofore  disagreed  as  to  the  true  meaning  of 
technical  terms  and  the  constitution  of  dynamical  quantities. 

The  prevalent  discordance  on  these  topics  has  caused  the  delay 
of  this  publication  for  over  ten  years,  in  which  interim  various 
discussions  thereupon  have  been  published,  both  in  Europe  and 
in  America,  indicating  the  confused  condition  of  the  subject. 

In  scientific  periodicals  we  rarely  find  a  sound  article  on  dy- 
namics, but  the  action  and  combination  of  physical  elements  are 
treated  as  if  governed  by  individual  judgment,  instead  of  the 
fixed  and  immutable  ordinances  of  nature. 

A  pamphlet  entitled  Principles  of  Dynamics,  exposing  the  con- 
fusion in  the  science  of  dynamics,  has  been  published  and  pre- 
sented to  the  principal  libraries  and  institutions  of  learning. 

In  addition  to  the  objects  above  mentioned,  this  treatise  con- 
tains sufficient  new  and  original  matter  to  warrant  its  publica- 
tion. 

The  Appendix  is  added  for  the  purpose  of  supplying  the  student 
with  materials  for  the  coming  revision  and  final  establishment 
of  an  international  system  of  metrology,  which  must  be  attended 

to  sooner  or  later. 

i«  5 


ALPHABETICAL  INDEX. 


A. 

PAGE 

Abrasion,  limit  of          .        .     88 
Aerostatics  and  dynamics      .     17 
Aggregate  forms,  three           .     99 
Air,  compression  of       .         .  266 
Air,  resistance  of  .         .         .167 
Air,    temperature    of    com- 
pressed     .         .         .         .267 
Animal  strength  or  capabil- 
ity     78 

Appendix  ....  307 
Arithmetics,  octonal  and  deci- 
mal ....  308 
Arithmetics,  duodenal  .  310 
Arithmetics,  senidenal  .  311 
Ascending  bodies  against 

gravity  ....  121 
Asteroids,  planetoids  .  .  287 
Astronomy  ....  281 
Atoms  .  .  .  .99 

Attraction  between  any  two. 

masses       .         .         .         .113 
Attraction,  maximum    .         .  295 
Attraction  in  and  above  the 
earth  .  104 


Attraction  on  the  earth's  sur- 
face   112 

Attraction  of  a  mountain       .  114 

B. 

Ballistic  pendulum  .  .  253 
Burden,  man  and  animal  .  78 

C. 

Camel,  capability  of  .78 

Calculus  applied  to  dynamics  154 
Catenary  .  .  .  21-23 
Centre  of  gravity  .  .  50 

Centre  of  gyration  .  .  184 
Centring  revolving  bodies  189,203 
Centrifugal  and  centripetal 

forces         ....  222 
Centrifugal  force  of  a  liquid     236 
Centrifugal    revolution   indi- 
cator         .         .         .         .235 
Changing  direction  of  a  mov- 
ing body   .         .         .         .220 
Circular  or  rotary  motion       .     76 
Circus  ring,  centrifugal  force  233 
Circular  saw          .         .         .79 


ALPHABETICAL    INDEX. 


Clock  dial,  duodenal  .  .  325 
Collision  or  impact  of  bodies  141 
Civilization  in  other  worlds  288 
Comets  .  .  .  .282 
Compass,  duodenal  .  .  327 
Conical  pendulum  .  .  246 
Connecting  rods  .  .  .  218 
Counting  beats  of  seconds  .  280 
Crank  motion  .  .  214,  217 
Cranes  .  .  .  48, 49 

Creation  of  worlds         .         .  281 

D. 

Decimal  arithmetic  and  me- 
trology     .        .        .         .308 
Decimal  system,  in  perfection 

of 308 

Density  and  volume  of  planets  302 
Diagrams  of  heavy  ordnance  257 
Differential  pulley  .  .  34 
Dredging  machines  .  .  80 
Duodenal  arithmetic  and  me- 
trology .  .  .  .311 
Dynamical  terms,  rejected  .  15 
Dynamics,  definition  of  .17,  57 
Dynamic  principles  .  .  57 
Dynamics  of  matter  .  .  149 
Dynamics  by  the  calculus  .  154 
Dynamics  of  heavy  ordnance  254 
Dynamics  of  sound  .  .  277 
Dvnamometers  .  269-276 


Earth,  the  planet  . 
Earthwork    . 


286 
42 


PAGE 

Effect,  definition  of  .64 

Elastic  and  hard  bodies  141-148 
Elements  and  functions  .  57 
Elephant,  capability  of  .  78 

Equilibrium  arid  stability  .  36 
Experiments  with  heavy  ord- 


.  256 


P. 


Falling  bodies        .         .         .117 
Falling  weight  by  rope  and 

pulley  .  .  .  .127 
Firing  a  ball  through  a  door  .  221 
Foot-pounds,  different  kinds  .  69 
Foot-tons  .  .  .  .68 
Flour  mills  .  .  .  .79 
Fly-wheels  .  .  .  198-202 
Force,  definition  of  .  17,  58,  59 
Force,  examples  .  .  .71 
Force  of  falling  bodies  .  .  133 
Force  of  inertia  .  .  .185 
Force  on  a  body  free  to  move  130 
Force  of  wind  .  .  .45 
Forces,  variable  .  .  .  157 
Friction,  work  of  .  .  .176 
Friction,  rubbing  .  .  .81 
Friction,  dynamometer  .  .  269 
Friction,  gearing  .  .  .91 
Friction,  screw-propellers  .  178 
Functions  and  elements  .  57 


Governors 
Gravitation 


229-233 
.     99 


ALPHABETICAL  INDEX. 


Gravity,  centre  of.  .  .50 
Gunpowder  pile-driver,Shaw's  263 
Gyration  .  .  .  .182 
Gyration,  centre  of  .  .  184 


Hammer,  steam     .         .  .135 

Hammer,  forge      .         .  .  213 

Heat  generated  by  light  .  285 

Heavy  ordnance    .         .  .  254 

Horse-mill    .                  .  .78 
Horse-power         .        .  63,  77-80 

Horse-power  of  machines  .     79 
Horse-power  of  steam-engines     77 

Hydrodynamics,  definition  of    17 

Hydrostatics,  definition  of  .     17 

I,  J. 

Impact  of  bodies  .         .  .  141 

Inhabitation  of  worlds  .  .  285 

Indicated  horse-power  .  .77 

Inclined  plane,  motion  on  .  125 

Index  of  contents           .  .       7 

Index  of  illustrations    .  .  330 
Inertia,  definition  of      .    115-185 

Inertia,  moment  of         .  .  184 

Indicator,  revolution      .  .  235 

Jupiter,  planet.      .         .  287 


Law  of  celestial  mechanics  .  289 
Levers,  different  kinds  of  26,  27 
Lever  of  a  force  .  .  .24 
Lifting  a  weight  vertically  .  172 


Light  generating  heat 
Llama  of  Peru 


PAGE 

285 
78 


Man-power   . 
Mars,  the  planet   . 
Mass,  definition  of 
Mass  of  the  earth 


63,64 

.  286 
.  106 
.  112 

Mass  of  the  earth  and  moon  .  299 
Mass  of  the  sun,  to  find  the  .  290 
Mass  of  any  planet,  to  find  the  300 
Mass  and  weight,  conversion 

of  ....  108-111 
Matter,  definition  of  .99 

Matt,  explanation  of  the  new 

term  .  .  .  .106 
Mean  force,  to  find  the  .  .  168 
Mechanics,  definition  of  .17 
Mercury,  the  planet  .  .  286 
Meteors  .  .  .  .282 
Metre,  folded  French  .  .  310 
Metrical  system,  imperfect  .  309 
Metrology,  octonal  .  .  308 
Metrology,  decimal  .  .  310 
Metrology,  duodenal  .  .  321 
Metrology,  senidenal  .  .  312 
Molecule  .  .  .  .99 
Momentum,  static  .  .  24 
Momentum,  dynamic  .  .  106 
Momentum,  stability  .  .  37 
Momentum  curves  .  .  219 
Moment  of  inertia  .  .184 
Motion,  different  kinds  of  .  59 
Motion,  straight  to  rotary  .  205 


10 


ALPHABETICAL  INDEX. 


Moon,  the  Satellite 
Music    . 

N. 

Nebulas 

Neptune,  the  planet 
Newton's  laws 
Noise,  dynamics  of 


PAGE 

.  286 
.  300 


281-283 
.  287 
.  289 
.  278 
Nomenclature  of  numbers     .  314 


O. 

Octonal  arithmetics  and  me-      « 

trology  .  .  .  .308 
Orbits  of  the  earth  and  moon  298 
Ordnance  dynamics  .  251-254 
Oxford  pendulum  .  247-249 

P. 

Parabolic  motion  of  projec- 
tiles ....  123,  168 
Parallelogram  of  forces  .  18 
Particle  of  matter  .  .  99 
Pendulum  ..  .  .  237-250 
Pendulum,  Oxford  diagrams  .  248 
Pendulum,  ballistic  .  .  253 
Percussion,  centre  of  .  .  250 
Pile-driver,  ordinary  .  .  135 
Pile-driver,gunpowder,Shaw's  263 
Planetary  system  .  .  283,  284 
Planets  in  our  -system  .  .  286 
Platform  scale  .  .  .32 
Power,  definition  of  .  58,  62 
Power  in  effects  or  foot- 
pounds .  .  .64,  73 
Power  of  a  horse  or  of  man  63,  78 


Projectile   thrown   up   verti- 
cally         .         .         .         .121 
Prong's  dynamometer   .         .  269 
Pulleys         .         .         .        34,35 


R. 

Radius  of  the  earth 
Radius  of  gyration 
Rails,  adhesion  on 
Recoil  of  ordnance 


.  226 
.  184 
.  94 
.  252 


Resistance  of  air  to  projec- 
tiles   164 

Resultant  of  forces  .  18-20 
Revolution  indicator,  Brown's  235 
Roads  .  .  .  .  78,  93 
Rolling  friction  .  .  .88 
Rolling-mills  .  .  79,  91 

S. 

Safety-valves,  graduation  of  .  31 
Saturn,  the  planet  .  .  287 
Saw-mills  .  .  .  .79 
Scales,  platform  and  weighing  32 
Screw-propellers,  friction  of  .  178 
Seconds,  counting  bea,ts  of  .  280 
Senidenal  arithmetics  .  .311 
Slip-railway  for  ships  .  .  90 
Slope,  natural  repose  .  .  42 
Sound,  dynamics  of  .  .  277 
Sound,  velocity  of  .  .  279 
Space  in  dynamics  .  58—65 
Spring,  elastic,  body  moving 

against  .  .  .  138,  174 
Stamp-mill  .  .  .  .210 
Stability  .  .  .  36-47 


ALPHABETICAL    INDEX. 


11 


PAGE 

Statics  .         .  17,  18 

Static  momentum          .  24,  33 

Steam-hammers     .         .  .  135 

Steamship  performance  .     81 

Systems  of  arithmetics .  .  312 

T. 

Threshing  machine  .  .     79 

Time,  definition  of  .  58,  62 

Time,  examples  of  .  .72 

Traction  on  roads  .  .     90 

U. 

Uranus,  the  planet        .        .  287 

V. 

Velocity,  definition  of  .  68,  60 

Velocity,  examples        .  .     72 

Venus,  the  planet          .  .  286 

Vis-viva,  explanation  of  .170 

Vis-viva  in  a  body  at  rest  .  171 
Volume  and  density  of  planets  302 

W. 

Walls,  retaining  .  .  38-44 
Weight,  explanation  of  .  105 
Waterworks.  ...  78 
Wind,  force  of  .  .  .45 
Work,  explanation  of  .  58, 67, 75 
Woric  of  a  spring  .  .174 
Work  of  friction  .  .  .  176 
Work,  diagrams  .  .  .216 
Work  of  mass  in  circular  mo- 
tion   182 

Workmanday        .  .     68 


Tables. 

PAGE 

Addition  and  multiplication  .  319 
Adhesion  on  rails  .  .  94 
Catenary  curve  .  .  .22 
Distance  traveled  by  sound  .  279 
Duodenal  and  decimal  num- 
bers   316 

Elements  of    the    planetary 

system  ....  284 
Elements  for  centre  of  gravity  52 
Experiments  with  heavy  ord- 
nance ....  256 
Falling  bodies,  velocity  of  .118 
Fly-wheels,  elements  of  .  212 
Friction,  sliding  .  .  .86 
Friction,  shaft  journals  .  .  87 
Friction,  governors  .  .  231 
Governors  .  .  .  .231 
Horse-power,  foreign  .  .  63 
Limit  of  abrasion  .  .  88 

Load  of  burden  .  .  .78 
Manual  and  animal  work  .  78 
Mass  converted  to  weight  .  110 
Natural  slope  of  granular  sub- 
stances .  .  .  .42 
Nomenclature  of  numbers  .  314 
Pendulum  oscillations  .  241,  245 
Pile-driver  performance, 

Shaw's  .  .  .  .268 
Steamship  performance  .  81 
Traction  on  roads  .  .  93 
Velocity  and  force  of  wind  .  45 
Velocity  of  the  earth  and  moon  299 
Weight  converted  into  mass  .  108 


INTRODUCTION. 


THIS  treatise  is  written  for  students  of  Mechanics,  and  the  technical 
terms  herein  adopted  are  those  used  in  the  machine-shop,  rejecting 
the  ideal  vocabulary  heretofore  used  in  textbooks  and  colleges. 

In  order  to  establish  a  standard  language  in  Mechanics,  it  is  hoped 
that  institutions  of  learning  will  approve  and  confirm  the  terms  and 
distinctions  of  dynamical  quantities  as  adopted  and  defined  in  these 
pages,  57  to  69,  inclusive. 

The  distinction  between  the  terms  force,  power  and  work  has  here- 
tofore not  been  clearly  defined,  but  either  of  these  terms  has  been 
promiscuously  applied  to  either  or  all  those  quantities,  according  to 
individual  caprices. 

Work  has  thus  been  distinguished  by  a  variety  of  terms  indicating 
different  characters  of  that  function.  It  has  generally  been  main- 
tained that  work  is  independent  of  time,  and  that  power  is  dependent 
on  time,  both  of  which  propositions  are  incorrect. 

REJECTED   TERMS. 

Energy.  This  term  is  used  to  denote  work,  but  the  sen.~e  of  it 
conveys  an  idea  of  a  different  virtue — namely,  that  of  activity  or 
vigor,  which  is  power.  We  say  that  a  man  has  a  great  deal  of  energy 
when  he  can  accomplish  much  Avork  in  a  short  time,  which  is  a  virtue 
of  power ;  but  if  he  accomplishes  the  same  quantity  of  work  in  a  much 
longer  time,  we  do  not  give  him  credit  for  much  energy.  The  term 
energy,  if  employed  at  all,  ought  to  be  applied  to  power  alone  ;  but  as 
we  have  the  expressive  term  power  for  that  function,  it  is  better  to  dis- 
pense with  the  term  energy  in  dynamics. 

The  term  work  is  the  proper  name  for  the  function  whi:ii  has  been 
called  energy. 


14  ELEMENTS  OF  MECHANICS. 

Quantity  of  Motion  is  a  term  also  used  to  denote  work,  which 
latter  is  a  different  function  from  that  of  motion.  The  sense  of  this 
term  is  inseparably  associated  with  an  idea  of  more  or  less  space, 
which  is  a  function  of  velocity  and  time  without  regard  to  force ;  and 
as  force  is  an  element  of  work,  the  term  quantity  of  motion  should  be 
rejected  as  improper  to  denote  that  function. 

The  words  Actual,  Total,  Quantity,  Mode,  Potential,  In- 
trinsic, Kinetic  Effort,  etc.  are  often  appended  to  terms  without 
affecting  the  nature  of  the  quantity  so  denoted  ;  the  objection  to  which 
is  that  one  and  the  same  quantity  is  differently  defined  according  to 
the  combination  of  these  appended  words. 

As  an  illustration  of  the  effect  of  these  appendages  to  terms  in 
dynamics,  we  may  apply  them  to  geometrical  quantities  ;  for  instance, 
volume  in  geometry  corresponds  to  work  in  dynamics,  and  may  be 
expressed  thus : 

Volume  of  a  cube. 

Cubical  volume  of  a  sphere. 

Total  intrinsic  volume  of  a  cone. 

Actual  potential  volume  of  a  cylinder. 

Total  intrinsic  quantity  of  volume  of  a  pyramid. 

The  actual  total  quantity  of  voluminous  cubic  inches  in  an  intrinsic 

cubic  foot  is  1728.     (A  cubic  foot  is  1728  cubic  inches.) 

All  these  expressions  mean  simply  volume;  as  the  different  com- 
binations of  terms  denoting  work  mean  simply  work,  or  the  product 
of  the  three  simple  elements  force,  velocity  and  time. 

The  rejection  of  the  superfluous  terms  will  render  the  subject  of 
dynamics  much  easier  to  teach,  learn  and  remember. 

There  is  also  an  expression  generally  used  in  the  English  language 
— namely,  "  Consumption  of  coal  per  horse-power  per  hour,"  which  is 
not  correct,  or  rather  it  is  nonsense.  The  intended  idea  should  be 
expressed,  "  Consumption  of  coal  per  hour  per  horse-power."  It  is  the 
fuel  which  is  divided  by  time,  and  not  the  power. 

The  consumption  of  fuel  is  work,  which  divided  by  time  is  power. 

There  exists  no  such  quantity  in  dynamics  as  power  per  time,  but 
power  multiplied  by  time  is  work,  and  work  per  time  is  power. 

The  following  list  shows  which  terms  are  herein  rejected  and 
adopted : 


INTROD  UCTION. 


15 


DYNAMICAL    TERMS. 


Rejected  Terms. 

Effort  of  force. 

Efficiency  of  force. 

Acting  force. 

Force  of  motion. 

Working  force. 

Quantity  of  moving  force. 

Quantity  of  motion. 

Mode  of  motion. 

Mode  of  force. 

Moment  of  activity. 

Mechanical  power. 

Mechanical  effect. 

Quantity  of  action. 

Efficiency. 

Rate  of  work. 

Dynamic  effect. 

Quantity  of  work. 

Actual  total  quantity  of  work. 

Total  amount  of  work. 

Actuated  work. 

Vis- viva. 

Living  force. 

Energy. 

Actual  energy. 

Potential  energy. 

Kinetic  energy. 

Energy  of  motion. 

Energy  of  force. 

Heat  a  form  of  energy. 

.Heat  a  mode  of  motion. 

Mechanical  potential  energy. 

Quantity  of  energy. 

Stored  energy. 

Intrinsic  energy. 

Total  actual  energy. 

Work  of  energy. 

Equation  of  energy. 

Equality  of  energy. 


Reason  for  Rejection. 

force. 


All  forces  act. 
Means  motive  force. 


Has  no  definite  meaning. 


Means  simply  power. 


Used  for  power  or  work. 
Means  simply  work. 


Formula  for  work. 
Primitive  and  realized  work. 


1020 


ELEMENTS  OF  MECHANICS. 


MECHANICS  is  that  branch  of  natural  philosophy  which  treats 
of  the  three  simple  physical  elements,  force,  motion  and  time ; 
with  their  combinations  constituting  power,  space  and  work. 

Mechanics  is  divided  into  two  distinct  parts — namely, 

STATICS  AND  DYNAMICS. 

STATICS  is  the  science  of  forces  in  equilibrium  or  at  rest :  it  is 
subdivided  into  three  branches,  treating  respectively  of  solids,  liquids 
and 


Statics,  strictly  speaking,  refers  to  forces  in  regard  to  solids. 
Hydrostatics  treats  of  the  pressure  and  equilibrium  of  liquids. 
Aerostatics  treats  of  the  pressure  and  equilibrium  of  air  or  gases. 

DYNAMICS  is  the  science  of  force  in  motion,  producing  power 
and  work ;  and  is  also  subdivided  into  three  branches,  embracing  re- 
spectively, solids,  liquids  and 


Dynamics,  strictly  speaking,  refers  to  power  and  work  of  solids. 
Hydrodynamics  treats  of  the  power  and  work  of  liquids. 
Aerodynamics  treats  of  the  power  and  work  of  air  or  gases. 


FORCE. 

The  term  Force  means  any  action  which  can  be  expressed  simply 
by  weight,  and  which  can  be  realized  only  by  an  equal  amount  of 
reaction.  Force  is  derived  from  a  great  variety  of  sources,  but 
whenever  it  is  simply  force  it  can  invariably  be  expressed  by  weight, 
without  regard  to  motion,  time,  power  or  work. 

A  detailed  explanation  of  force  is  given  in  Dynamics. 

2*  B  17 


18  ELEMENTS  OF  MECHANICS. 


STATICS. 


Statics  is  the  science  of  forces  in  equilibrium ;  it  embraces  the 
strength  of  materials,  of  bridges  and  of  girders;  the  stability  of 
walls,  steeples  and  towers ;  the  static  momentum  of  levers,  with  their 
combination  into  weighing  scales,  windlasses,  pulleys,  funicular  ma- 
chines, inclined  planes,  screws,  catenaria,  and  all  kinds  of  gearing. 

The  magnitude  and  direction  of  a  force  can  be  represented  by  a 
straight  line ;  but  no  force  can  be  realized  without  an  equal  amount 
of  resistance  in  the  opposite  direction,  which  likewise  can  be  repre- 
sented by  a  straight  line. 

PROBLEM  1. 

Let  the  line  F  represent  the  magnitude  and  direction  of  a 
force  acting  on  a  point  c  at  rest  or  in  motion ;  and  let  R 
represent  an  equal  amount  of  resistance  in  the  opposite  direc- 
tion ;  then  the  force  F  and  resistance  R  are  said  to  be  in 
equilibrium. 

When  two  or  more  forces  act  in  one  or  the  same  direction, 
as  in  the  case  of  several  men  pulling  on  one  rope,  the  several 
forces  are  added  together,  and  the  sum  is  considered  as  only 
one  force. 

PROBLEM  2. 

Let  the  lines  F  and  /represent -magnitudes  and  direc- 
tions of  two  forces  acting  at  a  point  c.  Complete  the 
parallelogram,  and  the  diagonal  R  will  represent  the 
magnitude  and  direction  of  the  consolidated  action  of 
the  two  forces  jPand/. 

The  diagonal  R  is  called  the  resultant  of  F  and  /. 
Make  R  equal  and  opposite  to  R;  then  R  represents  the 
magnitude  and  direction  of  the  resistance  at  the  point  c. 
The  lengths  of  the  lines  F,  f  and  R  in  any  unit  of 
measure  represent  the  corresponding  forces  expressed  in  any  unit  of 
weight. 

When  the  angle  between  the  forces  F  and  /  is  known,  the  magni- 
tude and  direction  of  the  resultant  R ,  or  resistance  R,  can  be  de- 
termined by  the  aid  of  trigonometry,  as  follows : 


STATICS. 


19 


Let  u  denote  the  angle  in  degrees  between  the  two  forces  F  and/, 
and  v  =  the  angle  between  .Fand  the  resultant  R ,  then  we  have 


tan.v  = 


f  sin.u 


F±fcos.u 
Resultant  12  =  sec.v(F  ±/  cos.u). 


.  2. 


Use  the  sign  +  when  u  is  less  than  90°,  and  -  when  u>90°. 

Example  1,  The  force  jP=68  and  /=42  pounds,  acting  at  the 
point  c,  Fig.  2,  at  an  Angle  «  =  65°  30'.  Required  the  force  and  direc- 
tion of  the  resultant  R  or  resistance  R. 


Formula  1,   ta#i.v 


42 


68  +  42  x  cos. 65° 


Fig.  3. 


which  is  the  tangent  for  24°  6',  the  direction  of  the  resultant  R  to  the 
force  F. 

Resultant  ^'  =  se<x24°  6'(68+42x<x>s.65°  30') -93.574  pounds,  the 
force  of  resistance  required, 

"  Nystrom's  Pocket-Book  "  contains  complete  tables  of  the  trigono- 
metrical functions,  both  natural  and  logarithmic. 

Natural.        Logarithm. 

swt.650  30'  =  0.90996  =  9.95902 
sec.24°  &   =L0955    =10.03961 

The  insertion  and  use  of  trigonometrical  functions 
in  algebraical  formulas  can  be  learned  without  being 
well  versed  in  the  science  of  trigonometry. 

PROBLEM  3. 

When  three  or  more  forces  are  acting  in  different 
directions  at  a  point  c,  Fig.  3,  find  first  the  resultant 
between  either  two  of  them,  say  F&ndf,  which  gives 
the  resultant  r  as  a  single  force.  Complete  the  par- 
allelogram r,  f,  and  the  diagonal  R  is  the  resultant 
of  the  three  forces  F,f,f. 

The  Formulas  1  and  2  are  applied  to  each  two 
forces  or  resultants,  as  in  Example  1. 

PROBLEM  4. 

Forces  may  be  applied  so  that  they  nearly  counter- 
balance one  another,  as  in  Fig.  4.  The  resultant  R  is 
smaller  than  either  of  the  three  forces  ^,/and/'. 


•20 


ELEMENTS  OF  MECHANICS. 


PROBLEM  5. 

'  To "  dissolve  a  force  into  two  component  parts,  Fig.  5.  Suspend  a 
rope  from  a  to  b,  and  hang  a  weight  F  on  it  at  c;  then  the  force  F  is 
dissolved  into  two  parts,  acting  one  on  the  line  c  a,  and  the  other  on 
c  b.  Draw  the  line  c  d  to  represent  the  magnitude  and  direction  of 
the  force  F.  Complete  the  parallelogram,  and  the  side  /  represents 
the  force  acting  on  c  d,  and  f  that  on  c  b. 

The  trigonometrical  expression  for  Fig.  5 
will  be : 

T-T    ri       •    ,        ^       •  n       Fsin.u 

F  :f  =sin..(u+v)  :  stn.u.     f  =  - —     — . 


Fsin.v 


:.sin.v.       f-- 


f  sin.u 


Fig.  6. 


PROBLEM    €. 

Let  a  rope  be  suspended  from  two  points  A  and  -3,  and  two 
weigh-ts  W  and  w  hung  upon  it.     The 
rope  is-  considered  to  have  no  weight, 
but  only  stretched  by  W  and  w.     Let 
j,.!/  F  denote  the  tension  of  the  rope  at  A, 

and  /=that  at  B',  then 


F:  W=  sin.u  :  sm.(u+x)  F- 


W  sin.u 


PROBLEM  7. 

Let  a  rope  be  suspended  from  the  points  A  and  B,  and  any  number 
of  weights  P,  Q,  R,  S,  Thung  upon  it.  Let  .F  denote  the  force  of  ten- 
sion at  A,  and  /=that  at  B.  Draw  the  lines  A  C  and  B  D,  so  that 
they  tangent  the  rope  at  A  and  B.  From  the  point  of  intersection 
draw  the  vertical  line  W  to  represent  the  sum  of  all  the  weights 


CATENARY. 


21 


P,  Q,  B,  S,  T=  W.   Complete  the  par- 
allelogram, of  which  the  side  F1  rep- 
resents the  force  of  tension  at  A,  and  * 
the  side /that  at  B. 

F:W= 


f:w  =  sin.v : sin.(x + v).  /= 


PROBLEM  8. 

Now  let  us  suppose  an  infinite  number  of  weights  to  be  suspended 
on  the  rope,  which  is  the  same 
as  to  consider  the  weight  of  the 
rope  itself,  or  that  of  a  chain. 
Draw  the  lines  and  complete  the 
parallelogram  as  described  in 
the  preceding  problem.  The 
vertical  diagonal  TF  represents 
the  weight  of  the  chain,  and  the 
forces  J^and/  are  as  in  the  pre- 
ceding formulas. 

PROBLEM  9.— THE  CATENARY. 

The  curve  formed  by  a  flexible  rope  or  chain  suspended  from  two ; 
points  is  called  a  catenary  or  chain-line. 

Let  v  denote  the  angle  of  the  curve  with  the  vertical  in  any  point  • 
P  whose  abscissa  is  x  and  ordinate  y ;  /  =  length  of  the  curve  0  P.  \ 
The  formulas  for  the  catenary  will  then  be 
lsin.*v 


1  sinh- 


,       ,         .  . 

hyp-h 

,  ., 

(cosec.v  —  1). 


sin.2v 

x  kyp.log.cot.^v 
"     (cosec.v  -I) 
jj  (cosec.v  -1) 

hyp.  log.  cot.  %v 


sin.v(l  -  sin.v) 


22 


ELEMENTS  OF  MECHANICS. 


The  formulas  for  the  catenary  are  very  difficult  to  manage,  on  ac- 
count of  the  angle  v  must  be  given ;  but  by  the  aid  of  the  following 
fcable  the  solution  becomes  very  simple  : 

Table  for  the  Catenary  Curve. 


Angle 

Abscissa 

Ordinate 

Curve 

7 

t. 

/ 

V. 

X. 

y- 

I. 

X 

y' 

30 

1.00000 

1.31690 

1.73210 

1.3169 

1.3153 

40 

0.55573 

1.01068 

1.19175 

1.8186 

1.1792 

45 

0.41421 

0.88137 

1.00000 

2.1278 

1.1346 

50 

0.30540 

0.76291 

0.83910 

2.4981 

1.1000 

54 

0.22078 

0.65284 

0.70021 

2.9570 

1.0725 

60 

0.15470 

0.54930 

0.5773o 

3^507 

1.0511 

62 

0.13257 

0.50940 

0.53171 

3.8425 

1.0438 

64 

0.11260 

0.47021 

0.48773 

4.1759 

1.0372 

66 

0.09484 

0.43169 

0.44523 

4.5518 

1.0314 

68 

0.07853 

0.39376 

0.40403 

5.0141 

1.0261 

70 

0.06418 

0.35637 

0.36397 

5.5527 

1.0213 

71 

0.05762 

0.33786 

0.34433 

5.8636 

1.0192 

72 

0.05146 

0.31946 

0.32492 

6.2079 

1.0171 

73 

0.04569 

0.30116 

0.30573 

6.5914 

1.0152 

74 

0.04030 

0.28296 

0.28675 

7.0213 

1.0134 

75 

0.03528 

0.26484 

0.26795 

7.5068 

1.0117 

76 

0.03061 

0.24681 

0.24933 

8.0631 

1.0102 

77 

0.02630 

0.22887 

0.23087 

8.7023 

1.0088 

78 

0.02234 

0.21099 

0.21256 

9.4445 

1.0073 

79 

0.01872 

0.19318 

0.19438 

10.820 

1.0062 

80 

0.01543 

0.17542 

0.17633 

11.372 

1.0052 

81 

0.01247 

0.15773 

OJ5838 

12.654 

1.0041 

82 

0.00983 

0.14008 

0.14054 

14.254 

1.0033 

83 

0.00751 

0.12248 

0.12278 

16.309 

1.0025 

84 

0.00551 

0.1  0491 

0.10510 

19.046 

1.0018 

85 

0.00382 

9.08738 

0.08749 

22.874 

L0013 

86 

0.00244 

0.06987 

0.06993 

28.613 

1.0008 

87 

0.00137 

0.05238 

0.05241 

38.171 

1.0005 

88 

0.00061 

0.03491 

0.03492 

57.279 

1.0002 

89 

0.00015 

0.01745 

0.01745 

114.586 

1.0000 

APPLICATION    OF   THE   CATENARY   TABLE. 

The  chain  for  a  suspension  bridge  of  300  feet  span  is  to  hang  60 

.feet  below  its  supports  on  the  piers.     The  chain  is  to  support  a  weight 

of  52,000  pounds,  uniformly  distributed  in  its  length.     Kequired  the 

length  of  the  chain  and  the  angle  v  and  strain  at  the  supports  ?    Half 

the  span  or  y  =  150  feet,  for  which  x  =  60  feet. 


60 


which  corresponds  nearly  with  an  angle  of  v  =  50°  in  the  table,  and 
the  length  of  half  the  chain  will  be  1  =  150  x  1.1  =  165  feet. 


CATENARY.  23 


The  strain  at  the  supports  will  be 

,,     TF«m.i7      52000  x  siw.500        .A,.n 
F=  -   —  <  --  -  =  40449  pounds. 

sin.Zv  «w.lOO° 

The  ordinates  x  and  y  and  the  length  I  for  any  angle  v  in  the  table 
are  found  as  follows  : 

When  v  =  50°  at  the  support,  find  x  and  y  where  v  =  70°  ? 

0.30540  :  0.06418  =  60  :  x.        x  =  0-06418x60  _  12.609  feet. 

0.30540 


0.76291  :  0.35637  =  150  :  y.        y  =    -  —  =  70.068  feet. 


Length  1  =  70.068  x  1.0213  =  71.56  feet. 

The  ordinates  can  thus  be  calculated  for  a  sufficient  number  of 
points  in  the  catenary  to  define  the  course  of  the  curve. 

The  strain  at  the  lowest  point,  or  centre  of  the  catenary,  will  be 
w  tan.v  =  26000  x  tan.  50°  =  30984  pounds  ; 

when  v  =  angle  at  the  piers,  and  w  =  half  the  weight  on  the  whole 
chain. 

The  catenary  is  not  a  line  of  the  conic  sections ;  its  figure  has  the 
appearance  of  a  parabola,  but  is  a  little  fuller  at  the  vertex. 

All  the  curves  of  the  conic  sections  are  of  the  second  order,  or  of 
the  exponent  n  =  2;  whilst  the  exponent  of  the  catenary  is  nearly 

The  formula  for  any  parabola  is  y  =  y'p  x, 
when  that  of  the  catenary  will  approach  y  =  y'p  x. 
Length  of  the  curve  OP  or  I  =  - 


PROBLEM  1O.  Fig'l°- 


To  find  the  resultant  of  two  parallel  forces 
F  and  W,  acting  at  the  ends  of  an  inflexible 
line  d,  e. 

Prolong  the  line  W  until  d  d'  is  equal  to 
F,  prolong  F  until  e  e'  is  equal  to  TF,  then 
join  d'  and  e',  which  will  cut  the  inflex- 
ible line  at  c.  Draw  from  c  the  resultant  R, 
equal  and  parallel  to  W+  F;  then  R  repre- 
sents the  magnitude  and  direction  of  the  re- 
sistance which  balances  the  two  forces  F 
and  W.  The  distance  a  =  d  c  is  the  lever  for 


ELEMENTS  OF  MECHANICS. 


the  force  W,  and  b  =  c  e  is  the  lever  for  F,  which  quantities  bear  the 
following  relation : 

F :  W=  a  :  b,  static  momentums  W  a  =  F  b. 


Fig.  11. 


STATIC    MOMENTUM. 

The  product  of  a  force  multiplied  by  its  lever,  is  called  static  'mo- 
mentum. The  resistance  at  c  is  called  the  fulcrum.  It  is  supposed 
in  this  case  that  the  forces  W  and  F  act  at  right  angles  to  the  levers 
a  and  b. 

The  lever  of  any  force  is  equal  to  the  right  angular  distance  from 
the  fulcrum  to  the  direction  of  that  force. 

When  the  static  momentums  Wa  and  Fb  are  equal,  then  the  forces 
TFand  F&re  in  equilibrium. 

PROBLEM  11. 

To  find  the  resultant  and  static  momentums  of  two  forces,  F  and 
W,  acting  obliquely  at  the  ends  of  an  inflexible  line,  d  e. 

Extend  the  lines  of  the 
forces  until  they  meet  at 
d'.  Move  the  forces  to 
d',  and  complete  the  par- 
allelogram as  shown  by 
Fig.  8 ;  then  the  diag- 
onal I?  is  the  resultant 
i  °^  ^e  ^wo  f°rcesi  the  con- 
tinuation  of  which  will 
cut  the  inflexible  line 
at  the  fulcrum  c ;  make 
R  =  jR',  the  resistance 
balancing  the  two  forces. 
The  levers  a  and  b  are 
drawn  from  the  fulcrum  c  at  right  angle  to  the  direction  of  the  re- 
spective forces. 

The  analogy  of  the  forces  and  levers  is  the  same  as  that  in  Fig.  10 — 
namely, 

F :  T7=  a  :  b,  static  momentums  Wa  =  Fb. 

The  resistance  It  is  less  than  the  sum  of  the  forces  F  +  W. 
Either  one  of  the  points  of  pressure,  d,  c  or  e,  may  be  considered  aa 
a  fulcrum,  as  represented  in  Figs.  12  and  13.- 


STATICS. 


25 


PROBLEM  12. 

Two    forces,   F  and    W,    acting  Fig.  12. 

against  one  another  on  the  line  c  d, 
of  which  c  is  the  fulcrum ;  to  find 
the  magnitude  and  direction  of  re- 
sistance, in  the  fulcrum  c. 

Prolong  the  forces  J^and  W  until 
they  meet  at  d',  from  which  set  off 
the  respective  forces  F'  and  W'\ 
complete  the  parallelogram  and  pro- 
long d'  c'  until  it  cut  the  line  d  e  in 
c,  which  is  the  fulcrum  for  the  forces, 
and  d'  c'  =  JR  is  the  resistance.  The 
levers  a  and  b  are  drawn  from  the 
fulcrum  c  at  right  angles  to  the  directions  of  the  respective  forces. 

F :  W=  a  :  b,  static  momentums  Wa  =  Fb. 


PROBLEM  13. 

Two  forces,  jPand  W,  acting  against  one  another  on  the  line  de,  of 
which  c  is  the  fulcrum,  to  find  the  magnitude  and  direction  of  re- 
sistance, and  the  fulcrum  c. 

Prolong  F  and    W  until  they  Fi&- 13- 

meet  at  d'.  Set  off  W  and  F 
from  d',  respectively;  complete 
the  parallelogram  and  continue 
d'  d  until  it  cut  the  line  e  d  at  c, 
which  will  be  the  fulcrum,  and 
d'c'  =  R  is  the  resistant  required. 

The  levers  a  and  b  are  drawn 
from  the  fulcrum  c  at  right  angles 
to  the  direction  of  the  respective 
forces. 

F :  W=  a  :  b,  static  momentums  Wa  =  Fb. 


ELEMENTS  OF  MECHANICS. 


LEVERS. 

Levers  are  classified  into  three  different  kinds  in  reference  to  the 
relative  position  of  the  force  F,  weight  TF,  and  fulcrum  c. 

When  the  fulcrum  c  is  between  the  force  F  and  weight  TF,  the 
lever  is  called  the  first  kind.  Fig.  14. 

When  the  weight  TFis  between  the  force  jPand  the  fulcrum  c,  the 
lever  is  of  the  second  kind.  Fig.  15. 

When  the  force  F  is  between  the  weight  TFand  the  fulcrum  c,  the 
lever  is  of  the  third  kind.  Fig.  16. 

The  two  forces  F  and  TFwill  be  distinguished  by  considering  F 
the  applied  force,  acting  on  its  lever  b,  to  lift  the  weight  TF,  acting  on 
its  lever  a. 

PROBLEM  14. 

ON  THE  ELEMENTS  OF  A  LEVER  OF  THE  FIRST  KIND. 

Rg.  14. 

T 

The  distance  between  TF  and 
F  is  denoted  by  /. 

F:W=a:b,      or      Fb=Wa. 


b  a  '  F  '  W 

Example  1.  Suppose  the  weight  TF=360  pounds  acting  on  a 
lever  a  =  2.5  feet,  what  force  Fis  required  on  a  lever  b  =  &  feet? 

„     TFa     360x2.5 

f  =  — —  = =  150  pounds,  the  answer. 

o  6 

That  is  to  say,  the  force  F=  150  pounds  acting  on  a  lever  5  =  6 
feet  will  just  balance  the  weight  TF=  360  pounds  on  the  lever  a  =  2.5 
feet,  without  regard  to  motion. 

Example  2.  The  force  .F=450  and  the  weight  TF=990  pounds 
acting  on  a  lever  a  =  3.6  feet,  required  the  lever  b,  for  the  force  F? 

,      TFa    990x3.6     , 

o  =  — —  =  — '- =  7.92  feet,  the  answer. 

When  the  force  ^and  weight  TF,  and  the  distance  I  between  them, 
are  given,  to  find  the  fulcrum  c  and  the  levers  a  and  b. 


LEVERS.  27 


The  lever  a  =  l-b,  and  the  lever  b  =  l-a. 

Fb=Wa,      or      F(l-a)=Wa.         Fl-Fa=Wa, 
-Fa=Wa-Fl,      or      Fa-Fl-Wa. 

Fa  +  Wa  =  Fl,     a(F+W)  =  Fl,      and      a=  -^—. 

F  +  W 

Example  3.  To  prove  the  correctness  of  the  formula,  let  us  assume 
the  same  value  of  the  quantities,  as  in  Example  2. 

I  =  3.6  +  7.92  =  11.52  feet.     Find  the  levers  a  and  b. 
F I  _ 450x11.52  _ 3_6          and  b  =  n_52_  3_6 


F+  W     4504-990 

PROBLEM  15. 

ON  THE  ELEMENTS  OF  A  LEVER  OF  THE  SECOND  KIND. 

F:W=a:b,       Fb 

F^^,       TF=^, 
b    '  a  ' 

Wa  Fb 


Example  4.  The  weight  TF=3696  pounds  acting  on  a  lever  a  =  0.5 
feet  ;  what  force  Fis  required  on  the  lever  b  =  15  feet? 

Wa    3696x0.5     . 

F=  -  =  --  =  123.2  pounds,  the  answer. 
b  15 

To  find  the  levers  and  fulcrum  when  the  weight  W,  force  F  and 
distance  I  are  given. 


Fb  =  Wa,       F(a  +  T)  =  Wa.         Fa+Fl=Wa, 
Fa-Wa=-Fl,       Wa-Fa  =  Fl. 

=  Fl,      and      a=  --. 


Example  5.    TF=6396,  JF=  150  pound,  and  the  distance  £=9  feet; 
required  the  levers  a  and  b. 

a=  Fl      =   15°x9    =0.2115655  feet  =  2.53877  inches. 
W-F    6396-15 

The  lever  b  =  a+J  =  9  +  0.2115655  =  9.2115655  feet. 


28  ELEMENTS  OF  MECHANICS. 

Example  6.  A  weight  TF=50  tons  is  to  be  lifted  by  a  force 
F=  0.25  tons,  acting  on  a  lever  b  =  45  inches ;  at  what  distance  shall 
the  weight  be  applied  from  the  fulcrum  c? 

Fb     0.25x45     _OOK.     ,        ,, 
a  = = — —  =  0.225  inches,  the  answer. 

F  and  W  can  be  expressed  in  any  unit  of  weight,  as  well  as  the 
levers  in  any  unit  of  length. 


PROBLEM  16. 

ON   THE   ELEMENTS   OF   A   LEVER   OF   THE   THIRD    KIND. 

Fig.  16. 

F:W=a:b,  or  Fb  =  Wa. 

?F  r  ^     Wa  W=F±_ 
a 

a 


Example  7.  A  weight  TF=50  kilograms,  acting  on  a  lever  a  =  4.5 
metres,  is  to  be  lifted  by  a  force  F=  88  kilograms ;  required  the 
lever  b ? 

,      Wa     50x4.5     0_Kh_ 

b  = = =  2.557  metres. 

F          88 

Given  the  force  F,  weight  W,  and  the  distance  I,  to  find  the  fulcrum 
and  levers  a  and  b. 

a  =  b  +  l,        b*=a-l. 
Fb  =  Wa,    or    F(a-l)  =  W  a. 
Fa-Fl=Wa,        Fa-Wa  =  Fl 
Fl 


a(F-  W)=  Fl,    and    a  = 


F-  W 


Example  8.    The  distance   Z  =  2.3   yards,    W=5  hundredweight, 
and  F=  5.5  hundredweight  ;  required  the  lever  a  ? 

Fl       5.5x2.3 


The  lever  b  will  be  25.3  -2.3  =  23  yards. 


LEVERS. 


29 


Fig.  17. 


PROBLEM  17. 

We  have  now  considered  the  levers  a  and  b  to  be  in  a  straight  line, 
on  which  the  forces  .Fand  Ware  acting  at  right  angles ;  but  in  prac- 
tice these  elements  occur  in  a  great  variety  of  relative  positions,  of 
which  one  is  illustrated  by  Fig.  17. 

The  apparent  lever  a  of  the  weight  inclines  under  the  fulcrum, 
whilst  that  of  the  force  F  rises 
above.  The  real  or  actual  levers 
for  these  forces  are  the  right- 
angular  distances  a  and  b ;  but 
even  these  are  not  in  a  straight 
line. 

Whatever  may  be  the  form  of 
the  apparent  levers,  or  whatever 
may  be  the  direction  of  the 
forces,  the  actual  levers  must  be 
considered  from  the  fulcrum,  at  right  angles  to  the  directions  of  the 
respective  forces.  The  static  momentums  of  equilibrium  and  the 
formulas  for  the  elements  are  precisely  the  same  as  that  explained 
for  the  three  kinds  of  simple  levers. 


Fig.  18. 


PROBLEM  18. 

We  have  heretofore  considered  the  levers  to  be  inflexible  lines  with- 
out weight,  which  will  answer  when  the  centre  of  gravity  of  the  mate- 
rial levers  is  in  the  fulcrum, 
like  that  of  a  weighing  bal- 
ance or  that  of  a  wheel ;  but 
this  centre  of  gravity  is  of- 
ten located  at  a  considerable 
distance  from  the  fulcrum,  as 
may  be  illustrated  by  Fig.  18, 
which  is  a  lever  of  the  first 

kind.     The  levers  a  and  b      

are  in  the  form  of  a  beam, 

resting  on  the  fulcrum  c,  and  its  centre  of  gravity  at  p.  Let  the 
weights  of  the  beam  be  represented  by  P,  acting  on  the  lever  x ;  there 
will  be  two  static  momentums,  F  b  and  P  x,  on  one  side  of  the  ful- 
crum, against  one,  W  a,  on  the  other  side. 


30 


ELEMENTS  OF  MECHANICS. 


Wa  =  Fb  +  Px, 
w  Fb  +  Px 

Fb  =  Wa-Px, 
„   Wa-Px 

Px^  Wa-Fb. 
D   Wa-Fb 

a 
Fb  +  Pa 

b    ' 
x   Wa-Px 

X 

Wa-Fb 

a"  w  ' 

F   ' 

P    ' 

Fig.  19. 


The  centre  of  gravity  p  may  be  found  experimentally  by  balancing 
the  beam  over  a  sharp  edge,  when  the  distance  x  can  be  measured 
from  the  fulcrum  c.  It  is  here  supposed  that  the  levers  a  and  b  are 
in  a  straight  and  horizontal  line. 

The  form  of  the  beam  affects  the  location  of  the  centre  of  gravity  p, 

but  when  this  centre  is  known 
the  shape  of  the  beam  does  not 
affect  the  static  momentums. 

The  pressure  on  the  fulcrum  c 
is  equal  to  W+P-F. 

PROBLEM  19. 

This  is  a  lever  of  the  second 
kind,  in  which  all  the  static 
momentums  are  on  one  side  of 
the  fulcrum. 


Px  =  Fb-  Wa. 

Fb  -  Wa 


Fb  =  Wa+Px, 
„     Wa  +  Px 

b        * 
Wa  +  Px 

a 
F^-Px 

F      ' 

Fig.  20. 

«  CL 

r_  n  

W      ' 
PROBLEM  20. 

.,  f         V  -       Pi         ' 

p= 


X 

Fb-  Wa 


This  is  a  lever  of  the  third 
kind,  representing  a  safety- 
valve  for  a  steam-boiler. 


Fb=  Wa  +  Px, 
Wa+Px 


p_ 


Fb  -  Wa. 
Fb-  Wa 


LEVERS.  31 


,      Wa+Px  Fb-Px  Fb-Wa 

0=-  -  •  a  =  -  x  —  - 

F  W  P 

The  formulas  are  the  same  as  those  for  the  lever  of  the  second  kind, 
Fig.  19. 

The  centre  of  gravity  of  the  lever  is  found  by  balancing  it  over  a 
sharp  edge,  and  the  distance  x  is  measured  from  the  fulcrum.  The 
weight  of  the  lever  is  found  by  weighing  it,  and  thus  the  momentum 
Px  is  obtained  simply  by  multiplying  the  weight  P  by  the  distance 
x,  which  is  a  constant  quantity  in  the  graduation  of  the  lever  for  dif- 
ferent pressures  of  steam. 

Example  9.  Suppose  the  safety-valve  to  be  three  inches  in  diam- 
eter, and  the  lever  to  be  graduated  between  pressures  of  20  and  50 
pounds  to  the  square  inch.  Area  of  the  valve  is  7  square  inches,  and 
7  x  50  =  350  pounds,  which  is  the  force  F.  The  lever  is  found  to 
weigh  P=10  pounds  and  #  =  18  inches,  which  momentum  Px  =  \Q 
x  18  =  180.  The  lever  b  =  4  inches  and  a  =48  inches  where  the 
weight  W  is  expected  to  indicate  a  steam  pressure  of  50  pounds  to 
the  square  inch.  Then  the  weight  will  be 

.    Fb-Px    350x4-10x18     ( 
W=  —        —  =  —  --  =  27.08  pounds. 

a  48 

Now  find  the  lever  a,  or  at  what  distance  from  the  fulcrum  shall 
the  weight  be  placed  for  balancing  a  steam  pressure  of  20  pounds  to 
the  square  inch.  F=  20  x  7  =  140  pounds,  and  the  lever 

Fb-Px     140x4-10x18 


Measure  off  this  distance  from  the  fulcrum  and  mark  it  with  the 
number  20.  Divide  the  distance  between  the  first  and  second  posi- 
tions of  the  weight  into  50  -  20  =  30  equal  parts,  and  number  them 
from  20  to  50.  The  lever  is  thus  graduated  for  the  required  pres- 
sures of  steam. 

For  greater  accuracy  it  is  necessary  to  weigh  the  valve  with  the 
lever  for  the  weight  P,  and  it  is  also  necessary  to  place  the  valve  on 
the  lever  at  its  proper  distance  from  the  fulcrum,  when  the  lever  is 
balanced  over  a  sharp  edge,  for  finding  its  centre  of  gravity. 


32 


ELEMENTS  OF  MECHANICS. 


PROBLEM  21. 

ON  THE  COMBINATION  OF  LEVERS,  AS   REPRESENTED  BY  Fig.  21. 

Fig.  21. 


Levers  can  be  combined  in  a  great 
variety  of  ways,  but  are  generally 
arranged  so  that  a  short  lever  acts  on 
a  long  one. 


W:  w  = 


W= 


:  a  a' a", 

w  b  b'  b"  , 

and 


or  Waa'a"  - 
Waa'a" 


wbb'b" 


a  a' a"  '  b  b' b" 

That  is  to  say,  the  big  weight  TPis  to  the  small  weight  w  as  the  prod- 
uct of  all  the  long  levers  b  is  to  the  product  of  all  the  short  levers  a. 

The  big  weight  multiplied  by  the  product  of  the  short  levers  is 
equal  to  the  small  weight  multiplied  by  the  product  of  the  long  levers. 

This  rule  will  hold  good  for  all  combinations  of  levers,  but  without 
considering  the  weight  and  centre  of  gravity  of  the  levers.  The  static 
momentum  of  the  weights  of  the  levers  is  determined  by  hanging 
small  weights  at  TFuntil  it  balances  the  levers  without  any  weight  at  w. 

PROBLEM  22.— PLATFORM   SCALE. 

This  figure  represents  a  platform  scale.  A  B  is  the  platform,  rest- 
Pig  22.  ing  on  levers  underneath. 

The  lower  figure  represents  the 
plan  of  the  levers,  with  the  platform 
removed. 

The  platform  rests  on  four  points, 
from  which  the  levers  extend  diag- 
onally to  the  centre,  and  there  are 
also  four  fulcrums,  all  of  which  only 
serve  to  give  parallel  motion  to  the 
platform,  upon  which  the  weight  W 
can  be  placed  at  one  corner,  or  be 
irregularly  distributed,  without  ef- 
fecting any  difference  in  the  weight 
w.  Although  the  platform  and  weight 
W  rest  upon  four  diagonal  levers,  the 
real  levers  of  the  static  momentums 
must  be  considered  parallel  to  the 


LEVEES. 


33 


sides  of  the  platform,  as  marked  on  the  figures,  and  only  one  of  them 
enters  into  the  formulas  and  calculations. 

W:w  =  bb'b":  a  a'  a",    or  Wa  a'  a"  =  w  b  Vb". 


. 
W 


wbb'  b" 


c*-   w  = 


Wa  a'  a" 


The  platform  and  levers  ought  to  be  so  arranged  and  proportioned 
that  they  balance  one  another,  or  that  they  are  in  equilibrium  with- 
out the  weight  W  and  iv. 


DIFFERENTIAL   BALANCE. 


Fig.  24  represents  a  convenient  balance-scale  for  weighing  heavy 
weights.  It  is  much  used  in  iron-foundries,  where  the  balance  with 
the  weight  is  hoisted  in  a  crane  for  weighing. 


L 

W:w  =  L:l, 

W-  W  ^ 

"~T' 
wi 

w 
The  lever  l  =  (a-b}. 


Wl 

L 

w  L 

~W' 


The  object  of  the  links  and  the  short  lever  is  to  bring  the  weight 
close  to  the  fulcrum,  or,  more  correctly,  to  obtain  a  short  lever  I. 

There  is  not  room  enough  on  the  main-balance  to  bring  the  direc- 
tion of  the  action  of  the  weight  W  sufficiently  close  k)  the  fulcrum  for 
weighing  heavy  weights.  The  levers  a  and  b  ought  to  be  equal  to  a' 
and  b' ,  respectively.  a  +  b  =  a'  +  b' ,  and  a  -  b  =  a'  -  b' . 

The  difference  between  a  and  b  is  generally  not  made  so  great  as 
shown  in  the  illustration.  For  a  lever  L  =  8  feet,  the  lever  I  is  only 
a  fraction  of  an  inch,  and  can  be  made  as  small  as  desired  by  making 
(a-b]  small.  The  scale  should  be  well  balanced  by  the  ball  B, 
without  the  weights  TFand  w. 


ELEMENTS  OF  MECHANICS. 


PULLEYS. 


Fig.  25. 


PROBLEM  23.— FIXED   PULLEYS. 

Let  a  force  F  be  applied  on  a  rope  Fig.  25,  extending 
over  a  fixed  pulley  R  R  to  lift  a  weight  W\  then 
F:  W=R:  R,  or  F  =  W. 

The  radii  of  the  pulley  act  as  levers  for  the  forces, 
and  as  the  radii  are  alike  on  both  sides,  the  force  F 
will  be  equal  to  the  weight  W. 

Should  the  forces  F  and  W  act  on  different  radii, 
Fig.  26,  then, 

F:W=r:R,         FR^Wr. 


Wr 
It  ' 

Wr 
F  ' 


W- 


FR 


FR 

W  ' 


Fig.  27. 


PROBLEM  24. 

DIFFERENTIAL   PULLEY-BLOCKS. 

The  endless  rope  or  chain,//  passes  over  a  pulley  with 
two  grooves  of  different  radii  R  and  r,  as  represented  by 
Fig.  27.  The  weight  W  is  equally  divided  on  the  ropes 
or  chains  //  which  tension  will  consequently  be  2/  =  W, 

The  momentums  on  each  side  of  the  centre  will  be 


R-r 


Example.  The  weight  TF=  1500  pounds,  the  radius  of 
the  large  sheave  is  R  =  6  inches,  and  the  small  sheave  r  =  5 
inches.  Required  the  force  F? 

„     1500(6-5) 

•^= 7T~^ —  =  125  Pounds- 


PULLEYS. 


35 


Fig.  28. 


PROBLEM  25.—  MOVABLE  PULLEYS. 

Single  movable  pulley,  Fig.  28, 
F:  W=l  :2. 


If  the  force  is  applied  at  a  and  acts  upward,  the 
result  will  be  the 


Fig.  31. 


Double  movable  pulleys,  Figs.  29  and  30, 
F:W=l:4:,         F=\W,         W=4F. 

The  force  F  pulls  on  only  one  rope, 
whilst  the  weight  W  hangs  on  four  ropes ; 
for  which  the  weight  is  four  times  the  force. 
For  three,  four,  etc.  movable  pulleys  there 
will  be  six,  eight,  etc.  ropes,  and  the  weight 
will  be  so  much  greater  than  the  force. 


PROBLEM  26.—  COMPOUND   PULLEYS. 

Let  n  denote  the  number  of  movable  pulleys,  then 


W 


Example.    What  weight  W  can  be  lifted  by  n  =  I 
or  four  movable  pulleys,  the  force  F=  300  pounds  ? 
W=  F2"  =  300  x23  =  300x8  =  2400    pounds,   the    an- 
swer. 

Fig.  31  represents  three,  or  n  =  3  movable  pulleys. 

MOTION    OF   THE   FORCE   AND   WEIGHT. 

Let  M  denote  the  motion  of  the  force  F,  and  ra  =  motion  of  the 
weight  W',  then 

M:m  =  W:F, 


Wm 


FM  =  Wm. 

FM 


F  '  W  ' 

This  proportion  will  hold  good  in  all  kinds  of  levers  and  pulleys. 
That  is  to  say,  the  motion  of  the  force  is  to  the  motion  of  the  weight  as 
the  weight  is  to  the  force. 


36  ELEMENTS  OF  MECHANICS. 


STABILITY  AND  EQUILIBRIUM. 

Stability  of  a  body  is  that  state  of  rest  which  cannot  be  disturbed 
by  an  infinitely  small  force. 

Equilibrium  of  bodies  is  that  state  of  rest  which  can  be  disturbed 
by  an  infinitely  small  force. 

Stability  and  equilibrium  are  generally  referred  to  the  force  of 
gravity  acting  vertically  or  at  right  angles  to  the  surface  of  the 
earth. 

The  force  with  which  a  body  rests  on  a  foundation  is  equal  to  the 
weight  of  the  body  acting  in  the  direction  of  the  vertical  line  passing 
through  its  centre  of  gravity.  Should  the  body  rest  on  only  one 
point  in  the  aforesaid  line,  it  will  be  in  equilibrium,  and  an  infinitely 
small  force  can  upset  it ;  the  body  has  therefore  no  stability. 

Let  the  body  rest  on  two  points  in  a  vertical  plane  passing  through 
its  centre  of  gravity,  and  the  two  points  of  support  are  one  on 
each  side  of  the  vertical  from  the  same  centre  of  gravity ;  then  the 
body  will  be  in  equilibrium  in  the  direction  at  right  angles  to  the  plane, 
and  stable  in  the-  direction  of  the  plane  ;  the  body  must  therefore  rest 
on  more  than  two  points  to  be  absolutely  stable. 


PROBLEM  27. 

Let  a  body  W  be  supported  on  a  tripod  or  three  points  a,  b,  d  in 
Fig.  32.  a  horizontal  plane.  Draw  from  the  centre  of  gravity 

of  the  system  the  vertical  line  c,  c' .  which  must 
fall  inside  of  the  triangle  a,  5,  d  to  make  the 
weight  and  tripod  stable,  and  an  appreciable  force 
is  required  to  upset  the  system. 

The  term  system  means  the  tripod  and  the 
weight,  or  any  number  of  bodies  rigid  together  and 
supported  at  the  points  a,  b,  c.  The  system  can  be 
upset  in  any  direction,  but  it  is  generally  considered 
to  be  upset  in  the  direction  over  either  one  of  the  sides  in  the  base- 
triangle,  which  is  called  the  line  of  fulcrum.  The  static  lever  upon 
which  the  system  acts  is  the  rectangular  distance  from  c'  to  the  line 
of  fulcrum. 

The  static  momentum  of  the  force  must  be  equal  to  the  momentum 
of  stability  to  bring  the  force  and  weight  in  equilibrium,  and  any 
additional  force,  however  small,  will  uppet  the  system. 


STABILITY  AND  EQUILIBRIUM. 


37 


MOMENTUM  OF  STABILITY. 

The  weight  of  the  system,  multiplied  by  its  static  lever,  is  called  the 
?nomentum  of  stability. 

The  height  of  the  centre  of  gravity  c  does  not  affect  the  momentum 
of  stability.  The  force  required  to  upset  the  system  depends  upon 
its  lever  of  action.  The  lever  of  the  force  is  the  rectangular  or 
shortest  distance  between  the  direction  of  the  force  and  the  line  of 
fulcrum. 


PROBLEM  28. 

A  body  W,  with  its  centre  of  gravity  at  c,  is  rest- 
ing on  the  plane  A  B.  A  force  F  is  applied  to  up- 
set the  body  over  the  fulcrum  at  A.  Continue  the 
direction  of  the  force,  and  draw  the  line  b  at  right 
angles  to  Ff  from  the  fulcrum,  then  b  is  the  lever 
for  the  force,  and  a  is  the  lever  for  the  weight  of 
the  body. 

F:W=a:  b,         Fb  =  W  a. 

,     Wa  .    Fb 


Fig.  33. 


b    ' 

Wa 
F  ' 


Fb 


Example.  The  weight  of  the  body  is  TF=3600  pounds,  and  its 
lever  a  =  1.8  feet. 

The  direction  of  the  force  F  is  such  that  its  lever  b  =  5  feet.  Re- 
quired the  force  F? 


„      Wa     3600x1.8     100fi  . 

F  = — =  1296  pounds. 

b  5 

Let  the  same  body  be  upset  over  the  fulcrum 
at  B,  the  lever  a  =  4  feet;  the  force  F  being  ap- 
plied so  that  the  lever  b  =  7  feet.  Required  the 
force  F? 

„    3600x4     om,n          , 
F=  — - —  =  2059  pounds. 


Fig.  34. 


276181 


38  ELEMENTS  OF  MECHANICS. 

PROBLEM  29. 

RETAINING   WALLS   FOR   WATER. 

Fig.  35  represents  a  wall  of  hydraulic  masonry  erected  for  re- 
taining water,  which  presses  on  the  inside  with  a  tendency  to  upset 
the  wall  over  the  fulcrum  at  A  . 

j-gi.  35.  For  simplicity  in  the  formulas  and  calcula- 

tions we  will  assume  a  definite  length  of  the 
wall,  say  one  foot.  The  pressure  of  the  water 
Hf  E!  against  the  wall  increases  with  the  depth  d, 
7  ^^^^EHj^rfrr  so  that  if  the  line  n  o  represents  the  water- 
pressure  per  unit  of  surface  at  the  bottom  of 
the  water,  then  any  other  horizontal  line  in 
the  triangle  m,  n,  o  represents  the  correspond- 
ing pressure  at  that  depth,  and  the  area  of 
the  triangle  represents  the  total  pressure 
against  the  wall.  The  centre  of  pressure  must  therefore  be  at  the 
same  height  as  that  of  the  centre  of  gravity  of  the  triangle,  which  is 
at  ^  d  from  the  bottom,  where  the  resultant  of  all  the  forces  acts  on  the 
lever  b  to  upset  the  wall. 

The  weight  of  a  cubic  foot  of  water  at  60°  Fahr.  is  62.33  pounds,  and 
the  total  pressure,  or  the  force  F,  acting  on  the  side  of  the  wall,  will  be 

F=  \  x  62.33  rf=  31.16  rf. 

The  lever  of  this  force-  is  5  =  £  d. 

The  static  momentum  of  the  force  will  be 


but  F=  31.16  d, 

or  the  static  momentum  =  |x  31.16  d*=  10.3866  d'1. 

The  weight  per  cubic  foot  of  hydraulic  masonry  may  be  assumed 
as  follows  : 

Brick,  90  pounds, 

Granite,  140  pounds. 

Let  A  denote  the  height  of  the  Avail,  t  =  its  thickness  in  feet,  and 
w  =  weight  per  cubic  foot  of  the  materials  in  the  wall. 
Then,  the  weight  of  the  wall  will  be 


when  the  momentum  of  stability  will  be,  without  water-pressure, 
Wa  =  htwa. 


STABILITY  AND  EQUILIBRIUM.  39 

From  this  momentum  of  stability  must  be  subtracted  the  static 
momentum  of  the  water-pressure,  and  the  remainder  will  be  the  real 
momentum  of  stability  of  the  wall  with  water-pressure. 

Let  W  denote  the  weight  of  the  real  momentum  of  stability,  or 
that  part  of  the  weight  of  the  wall  which  keeps  it  in  proper  position 
against  the  water-pressure  ;  then 


Example.  Suppose  the  wall  to  be  of  brick  masonry,  for  which 
U'  =  90,  the  height  7i  =  12  feet,  thickness  Z  =  2  feet,  and  the  depth 
of  water  d=W  feet.  Required  the  real  momentum  of  stability? 
The  lever  a  =  1  foot.  10.3866  x  102  =  1038.66  foot-pounds,  the  static 
momentum  of  the  water. 

12  x  2  x  90  x  1  =  2160  foot-pounds. 

Real  momentum  of  stability  =  2160  -  1038  =  1122  foot-pounds. 

That  is  to  say,  the  real  static  momentum  of  the  wall  is  slightly 
more  than  the  static  momentum  of  the  water-pressure,  or  that  the 
momentum  of  stability  of  the  wall  without  water-pressure  is  more 
than  double  the  static  momentum  of  the  water. 

This  rule  will  hold  good  for  any  length  of  wall  when  the  dimen- 
sions are  in  feet  and  the  pressures  in  pounds.  In  practice,  the  mo- 
mentum of  stability  of  the  wall  alone  should  be  at  least  four  times 
the  static  momentum  of  the  water-pressure. 

FROBLEM  3O.  Fig.  36. 

Let  the  materials  in  the  wall  in  the 
preceding  figure  and  example  be 
formed  into  a  wall  of  this  figure. 
That  is  to  say,  the  materials  in  the 
triangle  p,  q,  r  are  placed  at  the  base 
s,  A,  r,  and  formed  into  a  solid  wall 
A,  B,  q. 


The  static  momentum  of  the  water-pressure,  and  also  the  weight  of 
the  wall  W  will  be  the  same  as  in  the  preceding  example,  but  the 
base  A,  B  will  here  be  doubled,  or  t  =  4  feet,  instead  of  two.  The 
lever  of  the  weight  TFwill  be  a  =  \  x4  =  2.66  feet,  instead  of  one  foot, 
and  the  momentum  of  stability  will  be 

2160  x  2.66  =  5745.6  foot-pounds. 


40 


ELEMENTS  OF  MECHANICS. 


The  real  momentum  of  stab.  =  5745.6  - 1038.66  =  4707  foot-pounds, 
or  over  four  times  that  in  the  preceding  example,  where  the  wall  of 
the  same  amount  of  materials  was  vertical  on  the  outside. 

Now  let  the  same  wall,  with  the  same  materials  and  dimensions,  be 
turned  so  that  the  water-pressure  falls  on  the  inclined  side,  and  the 
fulcrum  of  the  wall  at  B. 

The  lever  of  the  weight  TTwill  now  be  only  £x4  =  1.33  feet,  and 
the  momentum  of  stability  without  water-pressure  will  be 

2160  x  1.33  =  2872.8  foot-pounds. 


Fig.  37. 


PROBLEM   31. 

The  water-pressures  act  at  right  angles  on  the 
inside  of  the  wall  A,  C,  the  centre  of  which  will 
be  at  £  of  the  depth  d,  or  at  £  of  A,  C  from  A. 
Continue  the  force  F,  and  draw  b  at  right  angles  to 
it  from  the  fulcrum  B.  With  the  assumed  dimen- 
sions of  the  wall,  12  feet  high  and  4  feet  base,  and 
depth  of  water  10  feet,  the  lever  b  will  be  1.2648, 
a  =  1.333  and  the  side  A,  C=  10.54  feet. 

The  force  F=  10.54  x  62.33  x  0.5  =  3284.79  pounds. 

Static  moment.  =  Fb  =  3284.79  x  1.2648  =  4154.6  foot-pounds. 

Real  moment,  stab.  =  4154.6  -  2872.8  =  1281.8  foot-pounds. 

This  proves  that  a  wall' of  the  assumed  dimensions  has  greater  sta- 
bility when  the  water  presses  on  the  vertical  side  than  when  on  the 
inclined  side. 

The  retaining  wall  and  its  inclined  side  may  be  so  proportioned 
that  the  static  momentum  of  the  water-pressure  increases  the  stability 
of  the  wall,  as  in  Fig.  38. 


PROBLEM:  32. 

ELEMENTS  OF  THE  RETAINING  WALL. 

Fig.  38. 

h 


tan.v  =  — . 
t 


tan.v 


t  tan.v. 


lit!  Moment,  stab.  =  Wa  =  \  hi  wx%  t  =  ±  hfw. 


STABILITY  AND  EQUILIBRIUM. 


ELEMENTS   OF   THE  WATER-PRESSURE. 

The  inclined  side  A,  C=d cosec.v. 
The  force  F=  \  d  cosec.v. 

The  distance  S  from  A  to  the  centre  of  pressure  will  be  8  =  £  d 
cosec.v. 

b  :  S  =  =p  t  =t  d  sec.v  :  d  sec.v. 
S  ( =f=  t  =t  8  sec.v)      =f  £    , 

0  = — ±  O  =  ^=  t  COS.V  ±  -i- 

5  sec.v  sec.-u 

j 

6  =  =p  t  cos.v  ± 


Subtract  the  smallest  term  from  the  largest,  and  the  remainder  is 
the  lever  b. 


When  t  cos.v  > 


3  sin.v 
the  fulcrum  B,  as  in  Fig.  37. 


,  the  direction  of  the  force 


of 


When  t  cos  .v  < 


d 


-,  the  direction  of  the  force  F  passes  inside  of 
3  sin.v 

fulcrum  B,  as  in  Fig.  38,  and  the  static  momentum  F  b  increases  the 
stability  of  the  wall. 

The  trigonometrical  functions  can  be  expressed  by  the  dimensions 
of  the  wall,  as  follows : 

E  =  length  of  the  hypotenuse  of  the  wall. 

h  t  E 

cos.v  =  — ,        sec.v  =  - — , 


t 

dE 


E 

cosec.v  =  — . 
h 


dR, 

:3A' 

PROBLEM  33. 

Whatever  may  be  the  shape  and  position 
of  a  retaining  wall,  its  momentum  of  stability 
is  W  a,  and  the  static  pressure  of  the  water 
acts  at  right  angles  to  the  surface  of  the 
wall. 

The  force  of  the  water-pressure  in  pounds 
is  equal  to  the  area  of  the  pressed  surface 
in  square  feet,  multiplied  by  half  the  depth 
in  feet  x  62.33. 


The  centre  of  pressure  is  at  one-third  of  the  depth  from  the  bottom. 


42 


ELEMENTS  OF  MECHANICS. 


PROBLEM  34. 

RETAINING   WALLS    FOR   EARTHWORK. 

- 40-  The  action  of  earth  or  other  granular  sub- 

stances, like  sand,  gravel,  grain,  etc.,  on  re- 
taining walls  is  the  same  as  that  described 
for  water,  and  the  momentums  are  calculated 
by  the  same  formulas,  with  the  only  excep- 
tion that  the  natural  slope  of  the  granular 
materials  diminishes  the  force  F  as  the  co- 
sine for  that  slope. 

The  natural  slope  of  a  granular  substance 
is  the  greatest  angle  with  the  horizon  at 
which  it  will  repose  in  a  heap.  Let  s  denote  the  angle  of  natural 
slope  and  W=  weight  per  cubic  foot  of  the  material  retained  by  the 
wall ;  which  values  for  some  substances  are  contained  in  the  following 
table : 

Natural  Slope  and  'Weight  of  Granular  Substances. 


Granular  Substances  Loosely  Heaped. 

S 
S. 

ope. 

cos.s. 

Weight, 
W. 

45 

070711 

Saw-dust,  wheat-flour  

44 

0.71934 

Broken  stone  or  coal  

43 

0.73135 

Malt-flour  

40 

0.76604 

Sand  (moist)  

39 

0.77715 

95 

Sand  (dry)  

38 

0.78801 

94 

Malt-corn  

37 

0.79863 

47 

Wheat,  rye  and  corn  
Peas  

36 
35 

0.80902 
0.81915 

45 

48 

Gravel  and  earth  

35  to  40 

0.8  to  0.75 

80  to  100 

The  force  F  per  foot  of  length  of  the  wall,  Fig.  40,  will  be 
F=  \w  d  cos.s. 

For  safe  calculation  we  may  assume  w  =  100  pounds  per  cubic  foot 
of  earth  or  gravel  pressing  against  the  wall,  and  s  =  32°  51'  the  safety 
angle  of  natural  slope,  which  cosine  is  0.84 ;  then  the  force  F  will  be 


fx  100  x  0.84  rf-42 


STABILITY  AND  EQUILIBRIUM.  43 

The  static  momentum  of  the  force  F  will  be  as  described  for  Fig. 
35  —  namely, 

Fb  =  \Fd,        but  F=42  d. 


The  momentum  of  stability  of  the  wall  alone  will  be 


of  which  the  weight  w  =  htw,  and  the  lever  a  =  %t,  when  the  sides 
of  the  wall  are  vertical. 


w  =  weight  per  cubic  foot  of  the  materials  in  the  wall,  which  is  90  for 
brick,  120  for  rubble  concrete  and  140  pounds  for  granite. 

Subtract  the  static  momentum  of  the  earth-pressure  from  the  mo- 
mentum of  stability  of  the  wall  alone,  and  the  remainder  will  be  the 
real  stability  of  the  wall. 

For  safety  in  practice  the  momentum  of  stability  of  the  wall  alone 
ought  to  be  at  least  four  times  the  static  momentum  of  the  earth-pres- 
sure, or 

or     hwf 


112  d*  10.583  d 

h  =  --          and£  =  —          -. 


"When  the  height  h  of  the  retaining  wall  is  equal  to  the  depth  d  of 
the  earth-pressure,  we  have 

w  V  =  112  d,  of  which  t  = 

Example.  The  depth  of  earth-pressure  is  d  =  2Q  feet,  to  be  retained 
by  a  vertical  granite  wall  of  w  =  140  pounds  to  the  cubic  foot.  Re- 
quired the  thickness  t  of  the  wall. 


This  example  is  for  a  wall  of  cut  granite  block  and  of  first-class 
workmanship. 

The  foundation  of  the  wall  should  be  much  thicker,  depending  on 
the  softness  of  the  ground  upon  which  it  stands. 

A  brick  wall  for  retaining  earth,  of  120  feet  high,  should  be  5  feet 
thick. 


44 


ELEMENTS  OF  MECHANICS. 


PROBLEM  35. 


Fig.  41. 


Walls  are  often  built  for  re- 
taining earth  of  greater  height 
than  that  of  the  wall,  as  illus- 
trated by  the  accompanying  fig- 
ure, in  which  case  the  centre  of 
pressure  of  the  earth  will  not  be 
at  one-third  from  the  bottom,  as 
in  the  former  cases. 

Let  e  denote  the  height  of 
centre  of  pressure  above  the  bot- 


tom of  the  wall;  then 


2d 
3  " 


to  the 


when  the  earth  above  the  wall  rises  with  its  natural 
height  d. 

Draw  the  force  F  at  right  angles  to  B  2  through  the  centre  of  pres- 
sure, and  continue  it  past  the  fulcrum  at  A,  as  shown  by  the  dotted 
line,  and  the  static  momentum  of  the  force  of  earth-pressure  can  thus 
be  found  graphically  on  the  drawing. 

The  mean  height  of  the  column  of  earth  pressing  against  the  wall 
is  d-  \  A,  and  the  area  of  the  base  of  that  column  is  the  line  .32x1 
foot,  and  the  force  F=  (B^w(d-\  K). 

The  centre  of  gravity  of  an  irregular  four-sided  section  of  the  wall 
is  found  as  follows : 

Divide  the  base  A,  B  and  the  top  1,  2  each  into  two  equal  parts, 
and  join  the  middle  points  7  and  8  with  the  opposite  corners,  as  shown 
by  the  figure  41 ;  divide  each  of  these  four  lines  into  three  equal 
parts,  join  3,  4  and  5,  6,  and  the  intersection  at  W  will  be  the  centre 
of  gravity  of  the  four-sided  section  of  the  wall,  and  the  lever  a  is  thus 
found. 

The  area  of  the  four-sided  section  is  equal  to  that  of  the  two  tri- 
angles A,  B,  7  and  1,  2,  8,  which,  multiplied  by  the  weight  w  per 
cubic  foot  of  the  material  in  the  wall,  will  be  the  weight  W. 

In  practice  the  momentum  of  stability  of  the  wall  alone  should  be 
at  least  four  times  the  static  momentum  of  earth-pressure,  or 


Fb. 


STABILITY  AND  EQUILIBRIUM. 


45 


PROBLEM  36. 

ON   THE   STABILITY   OF   TOWERS  TO  THE   FORCE   OF   WIND. 

The  momentum  of  stability  of  a  tower  or  any  other  structure  ex- 
posed to  the  wind  should  be  greater  than  the  static  momentum  of  the 
greatest  storm  to  which  the  object  may  be  exposed.  For  safety  in 
practice,  the  stability  should  be  at  least  four  times  the  momentum  of 
the  wind. 

The  static  momentum  is  equal  to  the  force  of  the  wind  acting  on 
the  object,  multiplied  by  the  height  of  the  centre  of  gravity  of  the 
surface  acted  upon. 

The  following  table  shows  the  force  of  wind  in  pounds  per  square 
foot  at  different  velocities : 

Table  of  Velocity  and  Force  of  "Wind. 


Miles 
per  hour. 

Feet  per  second. 

Force  per  sq. 
ft.-pound. 

Common  Appellation  of  the  Force  of  Wind. 

1 

1.47 

0.005 

Hardly  perceptible. 

2 
3 

2.93 
4.4 

0.020 
0.044 

I  Just  perceptible. 

4 

5.87 

0.079 

1 

5 
6 

7.33 

8.8 

0.123 

0.177 

j-  Gentle  pleasant  wind. 

7 

10.25 

0.241 

J 

8 

11.75 

0.315 

9 

13.2 

0.400 

10 

14.67 

0.492 

12 

17.6 

0.708 

Pleasant  brisk  gale. 

14 

20.5 

0.964 

15 

22.00 

1.107 

16 

23.45 

1.25 

j 

18 

26.4 

1.55 

1 

20 

29.34 

1.968 

[  Very  brisk. 

25 

36.67 

3.075 

J 

30 

44.01 

4.429 

] 

35 

51.34 

6.027 

[  High  wind. 

40 

58.68 

7.873 

j 

45 

66.01 

9.963 

} 

50 
55 

73.35 

80.7 

12.30 
14.9 

[  Very  high. 

60 
65 

88.02 
95.4 

17.71 

20.85 

I  Storm  or  tempest. 

70 

102.5 

24.1 

Great  storm. 

75 

80 

110. 
117.36 

27.7 
31.49 

>  Hurricane. 

100 

146.66 

50. 

Tornado,  tearing  up  trees,  etc. 

The  highest  known  wind  is  termed  "  tornado,"  which  moves  with 
a  velocity  of  100  miles  per  hour,  or  146  feet  per  second,  and  exerts  a 


46  ELEMENTS  OF  MECHANICS. 

pressure  of  50  pounds  per  square  foot  on  a  surface  stationed  at  right 
angles  to  the  direction  of  the  wind. 

Let  A  denote  the  area  in  square  feet  of  the  structure  exposed  to 
the  wind,  the  greatest  force  of  which  will  be  50  A  ;  b  =  height  of  the 
centre  of  gravity  of  the  exposed  surface  above  the  ground.  The 
greatest  static  momentum  of  the  wind  will  then  be  50  A  b,  and  the 
minimum  stability  of  the  structure  should  be  limited  to 


Fig.  42.  -pig.  42  represents  a  square  tower  with  parallel 

sides  exposed  to  a  tornado.     The  momentums  will 
then  be 

Stability,  Wa  =  Fb  static  momentum. 
Practically,  the  stability  Wa  ought  to  be  four 
times  the  static  momentum  Fb. 

Wa  =  4Fb,    or     Wa  =  20Q  Ab, 
200  A  b 


and  W= 

a 

in  which  W=  the  whole  weight  of  the  tower  in  pounds,  A  =  area  of 
one  side  of  the  tower  facing  the  wind. 

Example.  Suppose  the  tower  to  be  four  feet  square  and  20  feet 
high,  from  which  a  =  2  and  b  =  10  feet.  The  area  of  the  side  will  be 
A  =  4  x  20  =  80  square  feet.  What  weight  of  the  tower  is  required  to 
maintain  it  stable  to  a  tornado  ? 

_    200^6    200x80x10     . 

—  = =  80000  pounds. 

It  is  supposed  that  the  wind  acts  at  right  angles  on  one  side  of  the 
tower,  but  if  acting  in  the  direction  of  the  diagonal  of  the  square  sec- 
tion, a  greater  surface  will  be  exposed,  but  at  such  angle  to  the  wind 
that  the  acting  force  will  be  the  same  as  when  blowing  directly  on 
only  one  side. 

On  a  round  tower  of  diameter  equal  to  the  side  of  the  square  the 
force  of  the  wind  is  only  one-half  of  that  on  the  square  tower. 

PROBLEM  37. 

Fig.  43  represents  a  tower  or  chimney  of  the  form  of  a  conic 
frustum  of  diameters  d  at  the  top  and  D  at  the  base,  h  =  height  of 
the  tower;  all  in  feet. 

The  height  of  the  centre  of  gravity  b  is  calculated  from  the  fol- 
lowing formula : 


STABILITY.  47 


,     A    hi  D-d 

£\  

Example.  The  height  of  the  tower  A  =  260  feet,  diameter  at  the 
top  d  =  10  feet,  and  D  =  25  feet  at  the  base.  Required  the  height  of 
the  centre  of  gravity  of  the  surface  of  the  tower  ? 

9fiO      9fiO  /9r\     1 0\  Fi£- 43- 

,       ^DU       ^jDU/^O  —  J-V\       ..  ..  ^     .  rt    ., 


The  projecting  area  to  the  wind  is 

-  (Z>+ d)  =  —  (25  + 10)  =  4550  square  feet. 
2  2 


Of  this  area  only  one-half  is  effective  to  the  force 
of  the  wind,  or  A  •=  2275  square  feet.    The  static  momentum  of  a  tor- 
nado will  then  be 

^6  =  50  46  =  50  x  2275  x  111.43  =  12675162.5  foot-pounds. 

The  practical  momentum  of  stability  of  the  tower  should  be  at  least 
four  times  this  static  momentum. 

If  the  materials  in  the  tower  and  that  in  the  base  it  stands  on  were 
hard  enough  to  stand  the  crushing  force  at  the  fulcrum  in  upsetting 
the  tower,  the  lever  of  the  momentum  of  stability  would  be  half  the 
diameter  D  of  the  base ;  but  as  such  is  not  the  case,  in  practice  a  de- 
duction of  that  lever  must  be  made  in  accordance  with  the  softness  of 
the  materials  acted  upon  at  the  fulcrum. 

When  the  base  of  the  tower  is  square,  and  the  force  of  the  wind 
acts  at  right  angles  to  one  of  its  sides,  the  fulcrum  will  be  a  line ;  whilst 
on  a  circular  base  the  fulcrum  will  be  a  point  in  which  the  whole 
weight  of  the  tower  acts  to  crush  the  materials. 

In  ordinary  good  brickwork  the  lever  of  the  momentum  of  stability 
may  be  taken  at  0.7  of  the  radius  of  the  circular  base. 

The  weight  of  the  tower  in  the  preceding  example  should  then  be 

...    ±Fb    4x12675162.5     , 

W= = —  =  5794368  pounds. 

a  0.7x12.5 

A  small  deduction  ought  also  to  be  made  of  the  lever  in  a  square 
base,  where  it  may  be  taken  at  0.9  of  half  the  side  of  the  square. 

The  cohesive  force  of  the  materials  increases  the  stability  of  the 
structure  when  the  masonry  is  perfectly  solid  at  the  base. 


48 


ELEMENTS  OF  MECHANICS. 


CRANES. 

The  ordinary  crane  for  hoisting  purposes 
is  represented  by  Fig.  44. 

The  crane  is  held  in  position  or  supported 
by  the  shoe  a  and  cap  b.  The  dotted  lines 
represent  the  chain  or  rope  by  which  the 
weight  W  is  raised. 

The  chain  may  run  either  direct  from  the 
barrel  d  to  the  blocks  c  and/,  or  from  d  via 
e,  c  to/. 

The  body  of  the  crane  is  composed  of 
the  post  P,  gib  G  and  stay  8. 

The  weight  W  acts  on  the  lever  I,  and  the  reaction  of  force  F  in  the 
supports  a  and  b,  acts  on  the  lever  h. 
Static  momentums  F  h  =  W  I. 


A  •  r         F'        w 

These  are  the  principal  elements  of  the  crane. 
When  the  stay  8  is  in  the  direction  a,  c,  the  force  of  compression  8 
of  the  stay  will  be,  when  L  =  length  of  the  stay, 

WL 

h   ' 

Let  m  denote  the  distance  from  the  post  P  to  where  the  stay  8 
bears  the  gib  G,  and  n  =  distance  from  the  gib  G  to  where  the  stay  S 
is  supported  on  the  post  P.  Then  the  force  of  compression  of  the 

stay  will  be  /S  = . 

m  n 

Wl 

The  force  of  tension  t  on  the  gib  will  be  t  = . 

n 

The  forces  F,  8  and  t  must  be  expressed  by  the  same  units  as  that 
of  the  weight  W. 

The  weight  of  the  materials  in  the  crane  is  not  included  in  the 
formulas. 

The  tension  of  the  chain  is  found  by  the  formulas  for  pulleys. 

The  strain  of  the  stays  which  hold  the  cap  b  is  equal  to  the  force 
F,  or 


CEANES. 


49 


The  vertical  pressure  in  the  shoe  a  is  equal  to  the  sum  of  the 
weight  TTand  weight  of  the  crane. 

For  foundry  cranes  the  block  c  is  moved  in  or  out  to  suit  the  loca- 
tion of  the  weight  to  be  filled,  and  the  gib  Q  and  stay  8  are  both 
made  double,  so  that  the  chain  can  pass  between  the  parts. 

The  lateral  strain  z  on  the  gib  where  it  bears  on  the  stay  will  be 

Wl 


WHARF   CRANES. 

Wharf  cranes   for  loading  and   unloading  Fig.  45. 

boats  are  often  constructed  like  Fig.  45,  and 
for  which  the  static  elements  are  the  same  as 
those  for  the  foundry  crane.  When  the  di- 
rection of  the  chain  on  which  the  weight  W 
hangs  is  parallel  with  post  a,  b,  the  strain  F 
at  b  is  equal  to  the  horizontal  pressure  at  a. 

Static  momentums  F  h  =  W  I. 


The  lateral  strength  of  the  curved  part  of  the  post  must  compen- 
sate the  stay  in  the  foundry  crane. 


SHOP  CRANE. 

This  form  of  cranes  is  used  with  differential 
pulleys.  The  tension-rod  or  tie  T  serves  the  same 
purpose  as  the  stay  in  the  foundry  crane. 

L  =  length  of  the  tension-rod. 

T=  force  of  tension. 

F=  force  or  pressure  in  the  post  journals. 

/=  force  of  compression  of  the  gib. 
£=  distance  between  the  centres  of  the  block 
and  the  post. 

Static  momentums  F  h  —  W  I. 


Fig.  46. 


/. 

m  n  n  ' 


50  ELEMENTS  OF  MECHANICS. 


CENTRE  OF  GRAVITY. 

The  centre  of  gravity  of  a  body,  or  of  a  rigid  system  of  bodies,  is 
a  point  in  which,  if  there  suspended,  the  body  will  be  in  equilibrium 
in  any  position  it  may  be  placed,  like  that  of  a  wheel  or  circleplane 
suspended  in  the  centre. 

Fig.  47.  A   body  suspended   freely  from   any   point   a   will 

hang  with  its  centre  of  gravity  in  the  vertical  line  a  b. 
Now  suspend  the  body  from  another  point  c,  and  the 
centre  of  gravity  will  be  on  the  line  c  d  ;  then  when  the 
centre  of  gravity  is  on  both  the  lines  a  b  and  c  d,  it 
must  evidently  be  at  z,  where  the  two  lines  cross  one 
another. 

The  lines  a  b  and  c  d,  or  the  centre  of  gravity  z,  can 
also  be  found  by  balancing  the  body  on  a  sharp  edge. 

The  centre  of  gravity  of  any  figure  or  body  is  thus 
found  by  suspending  or  balancing  the  same  in  two  dif- 
ferent positions. 

CENTRE  OF  GRAVITY  OF  A  PARALLELOGRAM. 

The  centre  of  gravity  of  a  square,  rectangle  or  paral- 
lelogram is  the  point  where  the  two  diagonals  cross  one 
another. 

TO  FIND  THE  CENTRE  OF  GRAVITY  OF  A  TRIANGLE. 

Fig-  49-  Bisect  either  two  of  the  three  sides  of  the  triangle, 

and  draw  the  dotted  lines  to  the  opposite  angles,  as 
shown  by  Fig.  49.  The  crossing  of  these  lines  is  the 
centre  of  gravity  z,  which  is  one-third  of  the  dotted 
line  from  the  side. 

CENTRE  OF  GRAVITY  OF  A  SYSTEM  OF  TWO  BODIES. 

r'R-  -50-  Let  two  bodies  A  and  B  be  placed  at  a  dis- 

tance  c  between  their  centres  of  gravity,  and 
z  =  centre  of  gravity  of  the  system;  then 


A  a  and  B  b  are  moments  of  gravity,  which  are  alike  when  the 
system  is  supported  in  its  centre  of  gravity. 


CENTRE  OF  GRAVITY. 


51 


Then, 


£  c 


a  =  c—b         andi  =  c  —  a. 
=  B(c-a)  =  B  c-  B  a. 
c,     or 

,     , 
and     o  «= 


Ac 


A+B  A+£ 

Example.     A  =6  pounds,  B  =  10  pounds,  and  the  distance  c  =  18 
inches.     Required  the  distance  a? 

Be       10  x  18     . 

a  = = =  11.25  inches. 

A  +  B      6  +  10 

CENTRE   OF   GRAVITY 
of  a  system  of  any  number  of  bodies  placed  in  a  straight  line  o  p. 

Assume  any  point  o  from  which  to  refer  the  different  moments  of 
gravity,  Z=  A  +  B+C+D+E,  or  the  sum  of  the  weights  of  the  dif- 
ferent bodies ;  z  =  distance  from  o  to 
the  centre  of  gravity  of  the  system.  Flg-  51- 

Then  the  moments  of  gravity  are    . 


A,  B,  C,  etc.  are  weight,  and  a,  5,  €,  etc.  are  the  levers  of  the 
respective  momentums  of  gravity. 

CENTRE   OF   GRAVITY 

of  a  system  of  bodies  irregularly  placed,  with  the  centres  of  gravity  in 
one  plane. 

Draw  from  o  the  lines  op  and  o  q,  to  form  rectangular  co-ordinate 
axes  outside  of  the  system.  The  right-angular  moments  of  gravity 
are  referred  to  the  respective  axes,  as  will  be  understood  by  the  illus- 
tration . 


From  o  set  off  the   distance  z 
toward  p  and  draw  the  ordinate  /. 


Z  '  / 

which  gives  the  centre  of  gravity  of  the  system. 


52 


ELEMENTS  OF  MECHANICS. 


By  this  method  the  centre  of  gravity  of  a  great  variety  of  systems 
of  bodies  can  be  ascertained ;  for  instance,  that  of  a  vessel  or  steam- 
boat. The  axis  op  is  drawn  above  the  deck,  and  o  q  at  the  stern  of 
the  vessel.  A,  B,  C,  etc.  may  represent  weights  of  the  propeller, 
engine,  boiler,  coal  and  cargo.  The  centre  of  gravity  of  each  part  is 
ascertained  separately. 

When  a  steamer  is  constructed,  the  weight  and  moments  of  gravity 
of  each  part  ought  to  be  calculated  and  summed  up  to  correspond 
with  the  displacement,  for  which  the  following  form  of  table  is  set  up. 


ELEMENTS   OF   THE   STEAMER   SHOOTING   STAR. 

Horizontal  axis,  20  feet  above  load  water-line. 
Vertical  axis,  10  feet  aft  the  centre  of  rudder. 


Details. 

Weights 
of  the 
parts. 

Ho 
Lerer. 

Moments  c 

izontal. 
Moment. 

•f  g»Tfty. 

Vert 
Lever. 

ical. 
Moment. 

Rudder 

Tons. 
2.1 
12.8 
50.1 

48 
560 
500 
800 
3 

Feet. 
9 
47 
75 
110 
120 
50 
I5fr 
27.5 

Foot-tons. 
1&* 

601.6 
3757.5 
5280.0' 
76200,0 
25000J) 
12000fcfr 
8215 

Feet. 
26.5 
27.5 
21 
23 
25 
24 

se> 

IS 

Foot-tons. 
55.6 
3-52.6 
1052.1 
1104.0 
14000.0 
12000.0 
24000.0 
239.0 

Propeller  shaft  

Boilers  with  water  
Coal. 

Car<*o  aft 

Cargo  forward  

Propeller  

2102 

248940.5 

52803.3 

Horizontal  centre  of  gravity, 
centre  of  rudder. 


2102 


=  1 15.1  - 10  =  105.1  feet  from 


Vertical  centre  of  gravity, 


52803.3 
2102 


••  25.1  -  20  =  5.1  feet  below  load 


water-line. 

In  practice  the  calculation  is  carried  out  with  more  details. 

The  horizontal  position  of  the  centre  of  gravity  is  required  for 
knowing  how  the  vessel  will  float  in  regard  to  the  keel  and  load 
water-line. 

The  vertical  position  of  the  centre  of  gravity  is  required  for  de- 
termining the  stability  of  the  vessel. 


CENTRE  OF  GRAVITY. 


53 


TO  FIND   THE   CENTRE   OF   GRAVITY    OF   ANY    IRREGULAR    BODY 
RESTING   ON   TWO   SUPPORTS. 

Having  given  the  weights  p  and  Fis- 53- 

P  bearing  on  the  supports,  the  sum 
of  which  is  equal  to  the  weight  W 
of  the  body,  to  find  the  horizontal 
distance  z  of  the  centre  of  gravity 
from  the  support  p. 

W:P:d:z     and 


Pd 


W 

CENTRE  OF  GRAVITY  BY  THE  CALCULUS. 

By  the  aid  of  the  calculus  the  centre  of  gravity  of  regular  figures 
can  be  determined  with  precision. 

Any  figure  or  body  can  be  considered  as  a  system  of  bodies  com- 
posed of  its  parts.  The  sum  of  the  moments  of  gravity  of  all  the 
parts  is  equal  to  the  moment  of  gravity  of  the  body  or  system. 

Fig.  54  represents  a  triangle  with  the  base  B  and 
height  H.  Any  element  b  of  the  triangle  multiplied 
by  the  height  h  is  the  moment  of  gravity  of  that  ele- 
ment. 

Let  A  denote  the  area  of  the  triangle,  and  z  =  height 
of  centre  of  gravity  from  B.  Then  we  have 


Fig.  54. 


b:  B  =  (H-K): 


H 


and     b  = 

zA 


~  h 


H 

B(H-k) 


H 


zA  = 


B  H*    B  H' 


3 
B  H1 


(*-*)• 


That  is  to  say,  the  height  of  the  centre  of  gravity  from  the  base  is 
equal  to  one-third  the  height  of  the  triangle. 
5* 


54 


ELEMENTS  OF  MECHANICS. 


CENTRE   OF   GRAVITY   OF   A   QUADRANGLE. 

The  illustration  represents  a  quadrangle,  a,  c,  d,  e,  of  which  the  top  b 
is  parallel  with  the  base  B.     Prolong  the  sides  d  a  and  e  c  until  they 
meet  at/,  which  will  form  the  triangle  d,f,  e  of  height  H. 
h  =  height  of  the  quadrangle. 
A  =  area  of  the  quadrangle. 

=  area  of  the  triangle  a,f,  c. 
Q  =  area  of  the  whole  triangle  dtf,  e. 
Height  of  centre  gravity  of  the  whole  triangle  is 

- 

Height  of  centre   gravity  of  triangle   a,  /,  c  is 

-x)+h. 

z  =  height  of  centre  gravity  of  the  quadrangle. 


Moment  of  gravity,  QQff)  =  Az+  0[^(JI-  h)  +  A}.  .         .     1. 

K  ........    2. 


Height^", 


3. 


Area  of  quadrangle,    A=   (£+b).        .        .        .        .        .4. 


Area  of  triangle,  a,/  c,    0  =  -(H-K)  = 


A          t  i.  •      i      j  f 
Area  ot   triangle,  a,  j,  e, 


2        2(B-b) 
Moment  of  gravity,  Az  =  \Q  H-  \0  H- f  0  A.       ...     7. 

Insert  the  values  3,  4,  5  and  6  for  the  corresponding  quantities 
H,A,0  and  Q  in  Formula  7. 


= 


Bh 


Bh 


JB-b) 


Bh 
B-b 


Bh 


This   formula   reduces   itself  to     &— — —  ^— -,  which   gives  the 
height  of  centre  of  gravity  of  the  quadrangle  above  the  base  B. 


CENTRE  OF  GRAVITY. 


55 


CENTRE  OF  GRAVITY  OF  A  CONE. 


D  =  diameter  of  the  base  of  the  cone. 
h  =  height,  x  and  y  are  co-ordinates. 


Volume  of  cone,  C= 


12 


1G 


Fig.  56. 


12  16 

CENTRE   OF    GRAVITY    OF   A    PARABOLA. 

The  illustration  represents  a  parabolic  plane  of  abscissa  x  and  ordi- 
nate  y. 

z  =  height  of   centre   of  gravity   above   the 
base  B. 

d  =  depth  of  centre  of  gravity  from  the  ver-  j 
tex  o,  or  h  =  d+  z  and  z  =  h  —  d.  \ 

The  formula  for  a  parabola  is  y  =  2\/p  x.       -4 


I? 
The  parameter  of  a  parabola  is   />  =  —  -. 


The  area  of  a  parabola  is  A 

The  moment  of  gravity  from  the  vertex  is 


=  y  x  8x 


p  x 


-aVfv*- 


— 

yh 


- f 

,/iJ 


B  A" 


BJt 

2.5 


and   z  = 


That  is  to  say,  the  height  of  the  centre  of  gravity  of  a  parabolic 
plane  is  two-fifths  of  the  height  of  the  parabola. 


56  ELEMENTS  OF  MECHANICS. 

CENTRE   OF   GRAVITY   OF   A   PARABOLOID. 

Meaning  of  letters  is  the  same  as  in  the  preceding  example. 

Volume  of  a  paraboloid  (7=  —     —  . 
8 

Moment  of  gravity  for  solidity  of  the  paraboloid  is 


.  . 

4h  4h  12  h 

A        TT^A* 

=~'  *    '  Z  =  i 


The  centre  of  gravity  of  a  paraboloid  is  one-third  of  the  height 
from  the  base. 

EQUILIBRIUM   OF  STATIC   MOMENTUMS. 

The  illustration  represents  a  body  fixed  Figj>8. 

on  an  axis  0,  and  several  forces  A,  B,  C 
and  D,  JE,  J^are  acting  on  their  respective 
levers  a,  b,  c  and  d,  e,fto  rotate  the  body  in 
opposite  directions. 

Then  the  opposing  static  momentums  will  ^^—  ^"""^^SC 

be  in  equilibrium,  when  ^^^ 

A 


When  the  sum  of  any  number  of  static  momentums  acting  to  rotate 
a  body  in  one  direction,  is  equal  to  the  sum  of  the  momentums  acting 
in  opposite  directions,  then  the  opposing  forces  are  in  equilibrium,  and 
the  body  will  remain  at  rest. 

When  the  sums  of  the  opposing  static  momentums  are  not  alike,  the 
body  will  move  with  a  momentum  equal  to  the  difference  between 
the  two  sums. 


D  YNAMICS. 


57 


DYNAMICS. 


§  1.  Dynamics  is  that  branch  of  mechanics  which  treats  of  forces 
in  motion,  producing  power  and  work.  It  comprehends  the  action  of 
all  kinds  of  machinery,  manual  and  animal  labor  in  the  transforma- 
tion of  physical  work. 

Quantity  is  any  principle  or  magnitude  which  can  be  increased 
or  diminished  by  augmentation  or  abatement  of  homogeneous  parts, 
and  which  can  be  expressed  by  a  number. 

Element  is  an  essential  principle  which  cannot  be  resolved  into 
two  or  more  different  principles. 

Function  is  any  compound  result  or  product  of  two  or  more 
different  elements. 

A  function  is  resolved  by  dividing  it  with  one  or  more  of  its  ele- 
ments. 

Force,  Velocity  and  Time  are  simple  physical  elements. 

Power,  Space  and  Work  are  functions  of  those  elements. 

The  combinations  of  the  elements  in  the  functions  are  as  follows : 


FUNCTIONS. 

Power  P=F  V. 
Space  8  =  V  T. 


ELEMENTS. 

Force  =  F. 
Velocity  =  V. 
Time  =  T. 
Mass  =  M. 

F\M  =  V:  T. 

MOMENTUM. 

M  V=  F  T. 

These  are  the  fundamental  principles  in  dynamics. 


58  ELEMENTS  OF  MECHANICS. 


DYNAMICS  COMPARED  WITH  GEOMETRY. 

§  2.  In  geometry  we  have  three  fundamental  elements,  expressed  by 
the  terms  Length,  Breadth  and  Thickness,  which  serve  to  repre- 
sent to  the  mind  the  nature  of  the  several  properties  of  geometrical 
space. 

We  have  in  like  manner  in  dynamics  three  fundamental  elements, 
expressed  by  the  terms  Force,  Velocity  and  Time,  which  repre- 
sent the  nature  of  physical  Power,  Space  and  Work. 

Force  is  any  action  which  can  be  expressed  simply  by  weight 
without  regard  to  motion  or  time ;  it  is  an  essential  principle  which 
cannot  be  resolved  into  two  or  more  principles,  and  is  therefore  a 
simple  element.  Force  is  the  first  element  in  dynamics,  and  corre- 
sponds to  length  in  geometry. 

Velocity  is  speed  or  rate  of  motion,  an  essential  principle  which 
cannot  be  resolved  into  two  or  more  principles,  and  is  therefore  a 
simple  element.  Velocity  is  the  second  element  in  dynamics,  and 
corresponds  to  breadth  in  geometry. 

Time  is  duration,  or  that  measured  by  a  clock ;  it  is  an  essential 
principle  which  cannot  be  resolved  into  two  or  more  principles,  and 
is  therefore  a  simple  element.  Time  is  the  third  element  in  dynamics, 
and  corresponds  to  thickness  in  geometry. 

Power  is  the  product  of  the  first  and  second  elements,  force  and 
velocity,  and  is  therefore  a  function. 

Power  is  the  first  function  in  dynamics,  and  corresponds  to  the 
product  of  length  and  breadth,  which  is  surface  in  geometry. 

Space  is  the  product  of  the  second  and  third  elements,  velocity  and 
time,  and  is  therefore  a  function. 

Space  is  the  second  function  in  dynamics,  and  corresponds  to  the 
product  of  breadth  and  thickness,  which  is  the  area  of  a  cross-section 
of  a  solid  in  geometry. 

Work  is  the  product  of  the  three  simple  elements  force,  velocity 
and  time,  and  is  therefore  a  function. 

Work  is  the  third 'function  in  dynamics,  and  corresponds  to  the 
product  of  length,  breadth  and  thickness.which  is  volume  in  geometry. 

Work  is  also  the  product  of  the  element  force  and  function  space, 
because  the  function  space  contains  the  elements  velocity  and  time  : 
like  volume  in  geometry,  it  is  the  product  of  length  and  area  of  cross- 
section. 

Work  is  also  the  product  of  the  function  power  and  element  time, 
because  the  function  power  contains  the  elements  force  and  velocity  : 
like  volume  in  geometry,  it  is  the  product  of  area  and  thickness. 


MOTION.  59 


DETAILED   EXPLANATIONS   OF   THE    ELEMENTS  AND   FUNCTIONS. 

FORCE. 

§  3.  Force  is  any  action  which  can  be  expressed  simply  by  weight, 
and  is  distinguished  by  a  great  variety  of  terms,  such  as  attraction, 
repulsion,  gravity,  pressure,  tension,  compression,  cohesion,  adhesion,  re- 
sistance, inertia,  strain,  stress,  strength,  thrust,  burden,  load,  squeeze, 
pull,  push,  pinch,  punch,  etc.,  all  of  which  can  be  measured  or  ex- 
pressed by  weight  without  regard  to  motion,  time,  power,  or  work. 

All  bodies  in  nature  possess  the  incessant  virtues  of  attracting  and 
repelling  one  another,  which  action  is  recognized  as  force. 

The  physical  constitution  of  force  is  not  yet  known,  but  it  appears 
that  the  force  of  repulsion  is  derived  from  heat,  and  that  of  attraction 
from  cold  or  absence  of  heat. 

It  is  well  known  that  bodies  generally  expand  when  heated,  which 
proves  the  repulsive  force  of  heat;  and  that  bodies  contract  by  de- 
crease of  temperature  shows  the  superior  action  of  the  attractive  force 
in  the  absence  of  heat. 

The  two  forces  act  opposite  to  one  another,  and  when  the  force  of 
attraction  is  superior  to  that  of  repulsion,  the  body  will  maintain  the 
form  of  solid.  When  the  opposing  forces  are  equal,  the  body  will 
maintain  the  form  of  liquid,  and  when  the  force  of  repulsion  is  su- 
perior to  that  of  attraction,  the  body  will  have  the  form  of  a  gas. 

The  force  of  universal  attraction  will  be  explained  in  its  proper 
place. 

The  unit  for  measuring  force  is  any  assumed  weight,  as  pound  or 
ton. 

MOTION. 

§  4.  Motion  is  a  continuous  change  of  position  in  regard  to  as- 
sumed fixed  objects.  Motion  or  rest  are  only  relative ;  that  is  to  say, 
when  two  bodies  change  their  relative  positions,  either  one  of  them 
can  be  considered  at  rest  and  the  other  in  motion.  There  is  no  abso- 
lute rest  known  in  the  universe.  Our  earth  revolves  around  its  axis 
and  also  in  its  orbit  around  the  sun,  but  the  sun  also  revolves  both 
around  his  axis  and  in  a  small  irregular  orbit  in  regard  to  his  planet- 
ary system,  whilst  the  whole  planetary  system  moves  bodily  in  space. 

We  are  therefore  not  able  to  establish  any  absolute  rest,  but  must 
consider  motion  and  rest  only  as  relative. 

Motion  is  expressed  by  the  following  terms  :  Move,  going,  walking, 
passing,  transit,  involution  and  evolution,  run,  locomotion,  flux,  rolling, 
flow,  sweep,  wander,  shift,  flight,  current,  etc. 


60  ELEMENTS  OF  MECHANICS. 

Motion  of  Translation  is  when  a  body  moves  with  a  uniform 
velocity  in  a  straight  line  without  revolving,  or  when  each  particle 
of  the  body  moves  in  parallel  lines. 

Motion  of  Gyration  is  the  same  as  rotation,  or  when  all  the 
particles  in  a  body  describe  concentric  circles  around  one  common 
axis. 

Helicoidal  Motion  is  when  a  body  revolves  around  an  axis,  and 
at  the  same  time  moves  in  the  direction  of  that  axis,  like  a  screw 
or  rifle-ball,  which  is  the  result  of  the  two  motions  of  translation  and 
gyration. 

Lateral  Motion  is  that  motion  of  translation  which  a  body  gen- 
erates in  a  direction  out  of  its  greatest  extension.  Lateral  motion 
may  also  be  referred  to  a  stationary  line  or  plane  from  or  to  which 
a  body  is  moving.  Small  rivers  run  into  larger  ones  from  lateral 
directions. 

Rolling  Motion  is  the  combination  of  rotary  motion  of  a  body 
and  lateral  motion  of  the  axis  of  rotation,  like  a  wagon-wheel  rolling 
on  the  ground. 

There  are  also  different  kinds  of  motions,  designated  by  the  name 
of  the  path,  of  the  curve,  described  by  the  motion,  such  as  parabolic 
motion,  elliptic  motion,  cycloidal  motion,  etc. 

Space  Motion.  We  say  that  a  body  has  more  or  less  motion  in 
regard  to  greater  or  less  space  moved  through,  in  which  case  motion 
means  space. 

VELOCITY. 

§  5.  Velocity  is  rate  of  motion.  It  is  obtained  by  dissolving  the 
function  space  and  eliminating  the  element  time. 

8      V  T 
Velocity  =  —  = =  F",  the  element. 

Velocity  is  independent  of  space  and  time,  but  in  order  to  obtain 
its  value  or  expression  as  a  quantity,  we  compare  space  with  time. 
Thus,  when  the  value  of  velocity  of  a  moving  body  is  required,  we 
measure  a  space  which  the  body  passes  through,  and  divide  that  space 
with  the  time  of  passage,  and  the  quotient  is  the  velocity. 

Velocity  is  therefore  expressed  by  space  per  time,  as  feet  per  second 
or  miles  per  hour.  A  definite  velocity  can  be  expressed  by  any  units 
of  space  and  time,  because  velocity  is  an  essential  principle  which 
cannot  be  resolved  into  two  or  more  principles.  A  velocity  of  22  feet 
per  second  is  equal  to  a  velocity  of  15  miles  per  hour.  For  a  definite 
velocity  we  can  reduce  the  space  and  time  to  infinitely  small,  or  say 


VELOCITY. 


61 


absolutely  nothing,  without  affecting  the  velocity  in  the  least ;  which 
proves  that  velocity  is  an  independent  virtue  or  an  element. 

Absolute  Velocity  is  that  measured  or  observed  in  the  real  path 
of  motion. 

Apparent  Velocity  is  that  observed  from  an  assumed  fixed  point, 
and  which  can  be  measured  only  as  angular  velocity. 

The  terms  velocity  and  motion  are  often  used  for  one  another,  and 
neither  one  of  them  can  exist  without  the  other,  because  it  is  velocity 
which  generates  motion  by  the  aid  of  time. 

Suppose  a  body  to  move  from  JP  to  P' ,  an  ob- 
server at  a  would  see  less  velocity  or  motion 
than  would  one  at  b ;  but  both  motions  are  ap- 
parent velocities,  which  can  be  measured  only  by 
the  angles  P  a  P '  and  PbP',  whilst  the  abso- 
lute velocity  must  be  measured  in  the  path 
PP. 

Velocity  or  rate  of  motion  is  expressed  by  a  variety  of  terms,  as 
follows : 


QUICK  MOTION. 

I,  swiftness,  rapidity,  fleet- 
ness,  speediness,  quickness,  haste, 
hurry,  race,  forced  march,  gallop, 
trot,  run,  rush,  scud,  dash,  spring, 
etc. 


SLOW   MOTION. 

Slowness,  tardiness,  dilatoriness, 
slackness,  drawl,  retardation,  hob- 
bling, creeping,  lounging,  linger, 
sluggish,  crawl,  drawl,  loiter,  glide, 
langttid,  drowsy,  etc. 


Velocity  is  variously  expressed  by  different  units  of  length  and 
time  ;  for  example,  in  machinery  it  is  generally  estimated  in  feet  per 
second  or  minute  ;  in  steamboat  and  railway  traveling,  by  miles  per 
hour ;  the  velocity  of  light  and  electricity,  by  miles  per  second  ;  whilst 
the  retirement  of  the  Niagara  Falls  toward  Buffalo,  and  the  sinking 
and  rising  of  land  or  sea  may  fee  expressed7  in  feet  or  inches  per 
century. 

The  velocity  of  light  in  planetacy  space  is  about  200,000  miles  per 
second,  or  the  same  as  that  of  electricity  through  good  conductors. 
The  velocity  of  the  earth  in  its  orlit  around  the  sun  in  reference  to  a 
fixed  star  is  about  19  miles  per  second; 

The  velocity  of  a  point  on  the  earth's  equator  in  reference  to  the 
sun  is  about  1037  miles  per  hour,  or  1520  feet  per  second. 

Angular  velocity  is  an  apparent  motion  referred  to  a  fixed 
centre,  like  that  of  a  revolving  wheel  or  an  oscillating  pendulum ;  it 
is  measured  by  periodical  revolutions  or  oscillations,  generally  denoted 
bv  the  letter  n. 


62  ELEMENTS  OF  MECHANICS. 


TIME. 

§  6.  Time  implies  a  continuous  perception,  recognized  as  duration. 

Chronology  is  the  science  of  time. 

Instant  and  moment  are  points  of  time. 

Epoch  is  the  beginning  of  any  time  marked  with  some  remarkable 
events  and  recorded  by  historians  or  chronologists.  Era  is  nearly  the 
same  as  epoch,  except  that  it  is  generally  fixed  by  nations  or  denomi- 
nations, as  the  Christian  era. 

Time  is  expressed  by  a  great  variety  of  units — namely,  millennium, 
a  thousand  years ;  century,  one  hundred  years ;  score,  twenty  years ; 
year,  season,  month,  fortnight,  week,  day,  hour,  minute  and  second. 

Cycle  is  a  period  of  time  in  which  similar  phenomena  of  the 
heavenly  bodies  perpetually  occur,  such  as  the  cycle  of  the  sun,  a 
period  of  twenty-eight  years,  at  which  the  days  of  the  week  return 
to  the  same  dates  of  the  month. 

The  most  ordinary  unit  for  measuring  time  is  derived  from  the 
period  of  one  revolution  of  the  earth  around  its  axis,  in  reference  to 
the  sun,  which  is  called  a  solar  day,  and  divided  into  24  hours,  each 
hour  60  minutes  and  each  minute  60  seconds. 

Another  unit  of  time  is  the  period  occupied  by  the  earth  in  making 
one  revolution  around  the  sun,  in  reference  to  an  assumed  fixed  star, 
which  unit  is  called  a  sidereal  year,  and  contains  365  days,  6  hours, 
9  minutes  and  9.6  seconds,  mean  solar  time. 

We  have  no  positively  fixed  standard  for  measuring  time,  for  the 
period  of  one  revolution  of  the  earth  around  its  axis,  as  well  as  that 
around  the  sun,  are  both  liable  to  changes  by  meteors  or  heavenly 
bodies  falling  on  the  earth  which  accelerate  or  retard  its  motion.  Such 
changes  have,  within  the  time  of  our  astronomical  records,  been  so 
small  as  to  amount  to  probably  not  more  than  a  few  seconds  in  thou- 
sands of  years ;  but  it  may  happen  at  any  time  that  a  heavenly  body 
large  enough  might  strike  the  earth  and  cause  a  change  of  time  which 
would  at  once  be  perceived  by  us  all  without  the  aid  of  instruments 
for  that  purpose.  (See  Astronomy.} 

POWER. 

§  7.  Power  is  the  product  of  force  and  velocity ;  that  is  to  say,  a 
force  multiplied  by  the  velocity  with  which  it  is  acting,  is  the  power 
in  operation. 

The  English  unit  for  measuring  power  is  a  force  of  one  pound  acting 
with  a  velocity  of  one  foot  per  second,  and  called  one  foot-pound  of  power ; 
or  one  effect. 


POWER. 


63 


Man-power  is  a  unit  of  power  established  by  MORIN  to  be 
equivalent  to  50  foot-pounds  of  power,  or  50  effects ;  that  is  to  say,  a 
man  turning  a  crank  with  a  force  of  50  pounds  and  with  a  velocity 
of  one  foot  per  second  is  a  standard  man-power. 

An  ordinary  workingman  can  exert  this  power  eight  hours  per  day 
without  overstraining  himself. 

Horse-Power  is  a  unit  of  power  established  by  James  Watt,  to  be 
equivalent  to  a  force  of  33,000  pounds  acting  with  a  velocity  of  one 
foot  per  minute,  which  is  the  same  as  a  force  of  550  pounds  acting 
with  a  velocity  of  one  foot  per  second. 

That  is  to  say,  one  horse-power  is  550  foot-pounds  of  power  or 
effects,  or  11  man-power  of  50  effects  each. 

The  product  of  any  force  in  pounds  and  its  velocity  in  feet  per 
second,  divided  by  550,  gives  the  horse-power  in  operation. 

In  Watt's  rule  for  horse-power  is  given  a  velocity  of  only  one  foot 
per  minute,  which  is  equal  to  0.2  or  •J-th  of  an  inch  per  second — about 
the  velocity  of  a  snail.  The  force  corresponding  to  this  velocity  is 
33,000  pounds,  or  about  15  tons,  which  is  too  large  for  a  clear  con- 
ception of  its  magnitude,  and  a  horse  can  never  pull  with  such  a  force. 
A  horse  can  pull  550  pounds  with  a  velocity  of  one  foot  per  second, 
which  is  the  most  natural  expression  for  horse-power.  This  expres- 
sion is  used  on  the  continent  of  Europe. 


Foreign  Terms  and  Units  for  Horse-Power. 


Countries. 

Terms. 

Eng.  translation 

Units. 

Eng.  equivalent. 

English  

Horse-power. 

Horse-power. 

650  foot-pounds. 

550  foot-lbs. 

French  

Force  de  chevaL 

Force-horse. 

75  kilogr.  metres. 

542.47  foot-lbs. 

German  

Pferde-krafte. 

Horse-force.      •    513  Fuss-funde. 

582.25  foot-lbs. 

Swedish  

Hast-kraft. 

Horse-force.      |    600  skalpund-fot. 

542.06  foot-lbs. 

Russian  

Syl-lochad. 

Force-horse,      j    650  Fyt-funt. 

550  foot-lbs. 

The  quantities  Force  and  Power  are  clearly  distinguished  by 
different  terms  only  in  the  English  language. 

On  the  continent  of  Europe  horse-power  is  called  horse-force  or 
force-horse,  which  does  not  distinguish  force  from  power. 

Horse-force  can  be  considered  to  be  the  force  with  which  a  horse 
can  pull. 

The  word  force  is  needed  in  the  Swedish  and  German  languages. 

Puissance,  in  the  French  language,  means  power,  but  the  term  is 
not  generally  used  in  that  sense  in  dynamics. 

In  most  of  the  continental  languages  there  are  words  which  cor- 


64  ELEMENTS  OF  MECHANICS. 

respond  with  power,  but  they  are  not  used  in  that  sense  in  dynamics, 
where  the  term  force  is  used  for  power. 

The  Swedish  and  German  word  Jcraft  ought  to  be  used  for  power 
only,  and  not  for  force,  as  it  is  also  used. 

The  words  expressing  work  are  clear  and  definite  in  all  languages. 

EFFECT. 

The  term  effect  has  been  used  to  denote  both  foot-pounds  of  power 
and  foot-pounds  of  work.  These  two  kinds  of  foot-pounds  have  here- 
tofore not  been  clearly  distinguished  from  one  another,  for  which 
reason  the  term  effect  will  hereafter  be  used  only  to  denote  foot- 
pounds of  power. 

P  =  simple  power  in  foot-pounds  or  effects. 

F=  force  in  pounds. 

V  =  Velocity  in  feet  per  second. 

IP  =  Horse-power. 

Simple  power  P=F  V.     Effects. 

p 

Horse-power  H?  = 

550 

F  V 

Horse-power  H?  = . 

550 

Man-power         =  — . 

OU 

F  V 

Man-power          = . 

50 

One  horse-power  =  11  man-power. 

Any  action  of  force  producing  motion  is  power,  which  is  inde- 
pendent of  time. 

In  lifting  a  weight  vertically  the  force  F  is  equal  to  the  weight 
lifted,  but  in  drawing  a  load  on  a  road  the  force  of  traction  may  be 
considerably  less  than  the  weight  of  the  load. 

A  weight  of  1000  pounds  lifted  vertically  with  a  velocity  V=  2 
feet  per  second  requires  a  power  of  2000  effects,  which  is  equal  to 
2000  :  50  =  40  man-power,  or  2000 :  550  =  3.63  horse-power. 

A  load  of  1000  pounds  drawn  on  a  horizontal  road  with  a  velocity 
F=  2  feet  per  second  may  require  a  tractive  force  of  only  F=  100 
pounds,  and  the  power  will  be  200  effects,  or  4  man-power,  which  is 
only  one-tenth  of  the  power  required  in  lifting  the  same  weight  ver- 
tically. 


SPACE.  65 


Power  is  the  differential  of  work,  or  any  action  which  produces 
work,  whether  mental  or  physical. 

Power  multiplied  by  the  time  of  action  is  work — work  divided  by 
time  is  power.  Writers  on  dynamics  have  heretofore  assumed  that 
"power  is  the  work  done  in  a  unit  of  time,"  which  is  an  error. 

The  number  which  expresses  the  work  done  in  a  unit  of  time,  is 
equal  to  the  number  which  expresses  the  power  in  operation ;  but 
that  does  not  prove  the  two  quantities  to  be  alike. 

When  we  say  "  in  a  certain  time,"  which  is  equivalent  to  the  ex- 
pression "  per  unit  of  time,"  we  divide  by  the  time. 

Work  is  the  product  of  the  three  elements  Force,  Velocity  and 
Time,  and  when  we  say  "work  per  unit  of  time,"  we  eliminate  the 
time  from  the  work,  and  the  remainder  is  power,  which  is  the  product 
of  force  and  velocity. 

Power  may  be  expressed  by  the  following  terms : 

Traction,  propulsion,  impulsion,  capability,  puissance,  labor,  haul, 
drag,  draw,  heave,  occupation,  activity,  vigor,  energy,  etc.,  or  any  action 
which  implies  force  and  motion  without  regard  to  time. 

SPACE. 

§  8.  Space  in  dynamics  means  linear  space,  which  is  a  function  of 
the  second  and  third  elements,  velocity  V  and  time  T,  and  may  be 
likened  to  the  cross-section  of  a  solid,  which  is  a  function  of  breadth 
and  thickness. 

Space  is  herein  denoted  by 

8-  V  T, 

which  means  that  the  space  8,  expressed  in  linear  feet,  is  the  product 
obtained  by  multiplying  together  the  velocity  V  and  time  T. 

Space  cannot  be  generated  or  conceived  without  the  two  elements 
motion  and  time. 

In  viewing  a  short  linear  space  our  mind  flies  so  rapidly  over  it 
that  we  miss  the  conception  of  motion  and  time.  The  length  of  a 
piece  of  wood  has  been  generated  by  the  motion  and  time  required 
for  its  growth,  and  when  our  mind  surveys  that  length,  motion  and 
time  are  required  for  passing  from  one  end  of  it  to  the  other. 

In  like  manner,  when  we  at  a  glance  survey  a  bridge  we  are  unable 
to  appreciate  its  length  without  following  it  in  contemplation  from 
pier  to  pier,  with  some  expenditure  of  time  and  velocity  from  one  end 
to  the  other. 

So  also  when  we  imagine  objects  or  cities  far  distant  apart,  our  con- 
ceptions of  the  magnitude  of  the  distances  between  them  are  exces- 
6*  E 


66  ELEMENTS  OF  MECHANICS. 

sively  vague  and  inexact  without  an  imaginary  transit  over  them  from 
point  to  point,  which  .also  requires  both  time  and  velocity. 

The  distance  run  by  a  locomotive  or  steamboat  is  the  product  of  the 
velocity  and  time  of  the  trip,  and  no  distance  can  be  accomplished 
without  either  of  these  elements. 

Geometrical  spaces  are  magnitudes  of  three  different  kinds — namely, 
linear,  superficial  and  voluminous. 

Linear  space  is  that  generated  by  the  product  of  time  and  motion 
of  a  point. 

Superficial  space  is  that  generated  by  the  product  of  velocity  and 
time  of  lateral  motion  of  a  line. 

Voluminous  space  is  that  generated  by  the  product  of  velocity  and 
time  of  lateral  motion  of  a  plane. 

In  determining  velocity  it  appears  as  if  motion  were  dependent  on 
space  and  time,  because  we  measure  a  space  and  divide  it  by  the 
time,  in  order  to  form  a  conception  of  velocity  or  the  rate  of  motion. 
Space  is  the  product  of  time  and  velocity,  and  when  we  divide  that 
product  by  time,  the  quotient  will  be  the  simple  element  velocity  or 
rate  of  motion. 

In  other  words,  when  we  divide  the  space  with  time  we  resolve 
the  function  space  into  its  constituent  elements  and  eliminate  the 
time,  and  the  quotient  is  the  simple  element  velocity.  If  we 
divide  the  space  with  velocity,  the  latter  is  eliminated  from  the 
former,  and  the  quotient  is  the  simple  element  time. 

Space  in  dynamics  means  the  generation  of  that  space  by  velocity 
and  time.  A  line  of  any  kind  cannot  be  drawn  without  velocity  and 
time. 

A  locomotive  running  with  a  uniform  velocity  of  30  miles  per  hour 
will  make  2640  feet  per  minute  or  44  feet  per  second;  and  if  we 
diminish  the  space  and  time  to  infinitely  small,  or,  say,  absolutely 
nothing,  the  velocity  of  30  miles  per  hour  is  still  constant  when 
passing  that  time  and  space  reduced  to  a  point. 

In  geometry  length  is  an  element  without  regard  to  velocity  or 
time,  but  in  dynamics  linear  space  means  a  physical  function  gene- 
rated by  velocity  and  time. 

Length  is  a  geometrical  element. 

Space  is  a  physical  function. 

In  regard  to  space  being  composed  of  velocity  and  time,  the  follow- 
ing question  has  been  asked :  The  distance  to  the  moon  is  space  ;  what 
has  that  to  do  with  velocity  and  time  ?  The  moon  has  never  been  on 
the  earth,  and  consequently  not  moved  from  it  with  velocity  and 
time  ?  The  answer  is  as  follows :  In  order  to  find  the  distance  to  the 


WORK.  67 


moon,  a  space  is  measured  on  the  earth's  surface,  and  is  obtained  by 
velocity  and  time,  which  is  converted  into  a  base-line — namely,  the 
diameter  of  the  earth.  The  moon's  horizontal  parallax  is  next 
measured,  through  which  the  velocity  and  time  in  the  base-line  is 
multiplied  until  it  reaches  the  moon.  Thus,  the  space  to  the  moon  is 
composed  of  velocity  and  time. 

WORK. 

§  9.  "Work  is  the  product  obtained  by  multiplying  together  the 
three  elements,  force  F,  velocity  Fand  time  T,  or  work  7T=  F  V  T. 

"Work  may  also  be  expressed  by  K=  F  S,  or  the  product  obtained 
by  multiplying  together  the  force  F  and  space  S,  in  which  it  appears 
as  if  work  was  independent  of  time ;  but  the  time  is  included  in  the 
space  S  =  V  T.  A  given  amount  of  work  may  be  performed  in  any 
desired  length  of  time,  but  the  work  is  nevertheless  dependent  on 
whatever  time  consumed  in  its  execution. 

A  definite  quantity  of  work  is  not  confined  to  any  definite  ratio  or 
relation  to  either  of  its  constituent  elements,  for  either  one  or  two  of 
them  may  vary  ad  libitum,  but  only  at  the  expense  of  the  remaining 
two  or  one.  A  definite  quantity  at  work,  only  requires  a  definite 
product  of  the  combined  actions  of  the  three  elements.  Work  is  thus 
dependent  on  time  as  well  as  on  force  and  velocity,  for  without  either 
one  of  these  three  elements  it  ceases  to  be  work. 

If  work  was  independent  of  time,  then  any  amount  of  work  could 
be  accomplished  in  no  time. 

The  greatest  amount  of  work  known  to  have  been  accomplished 
in  the  shortest  time,  is  that  in  the  explosion  of  nitro-glycerine,  which 
is  instantaneous  to  our  perception ;  but  it  required  time  notwith- 
standing. 

Work  may  also  be  expressed  by  K=  P  T,  or  the  product  of 
power  and  time. 

The  work  of  a  steam-engine  operating  with  a  constant  power,  will 
be  directly  as  the  time  of  operation,  and  so  with  all  labor,  whether  it 
be  mechanical  or  manual. 

The  longer  we  toil  the  more  work  will  be  done,  but  if  we  have  no 
time  to  do  the  work  it  will  remain  undone. 

Much  of  the  confusion  in  dynamics  has  arisen  from  misconception 
of  the  difference  between  specific  quantities  and  abstract  numbers. 

When  a  quantity  is  multiplied  by  an  abstract  number,  the  product 
will  be  the  same  as  the  sum  of  so  many  concrete  quantities  added 
together  as  indicated  by  the  number,  and  the  operation  will  change 
the  magnitude,  but  not  the  nature,  of  that  quantity.  But  when  a 


68  ELEMENTS  OF  MECHANICS. 

quantity  is  multiplied  by  another  quantity,  the  product  will  be  a 
third  quantity  of  different  nature  from  that  of  its  constituent  quan- 
tities. 

A  force  of  2  pounds  working  with  a  velocity  of  3  feet  per  second  is 
a  power  of  6  foot-pounds,  which,  multiplied  by  the  abstract  number  4, 
will  be  a  power  of  24  foot-pounds.  But  the  same  power,  6  foot-pounds, 
multiplied  by  the  quantity  4  seconds,  will  be  24  foot-pounds  of  work. 

Three  square  feet  multiplied  by  the  abstract  number  2  will  be  6 
square  feet,  but  when  three  square  feet  are  multiplied  by  a  thickness 
of  2  feet  the  product  will  be  6  cubic  feet. 

The  erroneous  expression  that  "  power  is  the  work  done  in  a  unit 
of  time  "  implies  that  power  is  a  portion  of  work.  Power  multiplied 
by  a  unit  of  time  is  work,  and  work  divided  by  a  unit  of  time  is 
power.  In  both  these  cases  the  unit  of  time  does  not  change  the  nu- 
merical value  of  the  quantities,  but  it  changes  their  nature  from  one 
to  the  other. 

A  pool  of  water,  say  1000  square  feet  of  surface,  is  frozen  over 
with  ice  one  foot  thick,  and  there  will  consequently  be  1000  cubic 
feet  of  ice  in  the  pool,  which  though  identical  in  number  with  that 
of  the  surface,  does  not  prove  that  a  square  foot  is  a  cubic  foot. 

The  surface  of  the  pool  represents  power;  the  thickness  of  the  ice 
represents  time,  and  the  volume  of  the  ice  represents  the  work  con- 
sumed in  freezing  it. 

UNITS   OF  WORK.— FOOT-POUND. 

The  English  unit  of  work  is  assumed  to  be  that  accomplished  by  a 
force  of  one  pound  raising  an  equal  weight  one  foot  high,  which  unit 
is  called  a  foot-pound.  Then  a  force  of  6  pounds  working  through 
a  space  of  4  feet  is  equivalent  to  24  foot-pounds  of  work. 

This  unit  is  very  convenient  for  small  amounts  of  work,  but  it  is 
too  small  for  many  purposes  in  practice. 

FOOT-TON. 

English  ordnance  officers  have  adopted  a  larger  unit  for  work, 
namely,  foot-ton,  which  is  used  for  expressing  work  of  heavy  ord- 
nance. It  means  the  work  of  lifting  one  ton  one  foot  high. 

WORKMANDAY. 

A  laborer  working  eight  hours  per  day  can  exert  a  power  of  50 
foot-pounds.  A  day's  work  will  then  be  50x8x60x60  =  1,440,000 
foot-pounds  of  work,  which  may  be  termed  a  workmanday. 


FOOT-POUNDS.  69 


All  kinds  of  heavy  work  can  be  estimated  in  workmandays,  such 
as  the  building  of  a  house,  a  bridge,  a  steamboat,  canal  and  railroad 
excavations  and  embankments,  loading  or  unloading  a  ship,  powder  and 
steam-boiler  explosions,  and  the  capability  of  heavy  ordnance,  etc. 

The  magnitude  of  the  unit  workmanday  is  easily  conceived,  be- 
cause it  is  that  amount  of  work  which  a  laborer  can  accomplish  in  one 
day.  Work  expressed  in  foot-pounds,  divided  by  1,440,000,  gives  the 
work  in  workmandays. 

A  work  of  20  workmandays  can  be  accomplished  by  20  men  in  one 
day,  by  one  man  in  20  days,  by  4  men  in  5  days,  or  by  10  men  in  two 


•<..  10.     DIFFERENT   KINDS   OF   FOOT-POUNDS. 

There  are  four  different  kinds  of  foot-pounds  in  mechanics — namely, 

1st.  Foot-pounds  of  static  momentum,  which  are  force  in  pounds 
multiplied  by  its  lever  of  action  in  feet. 

2d.  Foot-pounds  of  dynamic  momentum  are  mass  expressed  in 
pounds,  multiplied  by  velocity  in  feet  per  second. 

3d.  Foot-pounds  of  power  (effects)  are  force  in  pounds,  multiplied 
by  velocity  in  feet  per  second. 

4th.  Foot-pounds  of  work  are  force  in  pounds,  multiplied  by  space 
in  feet. 

It  will  be  observed  that  foot-pounds  of  static  momentum  and  foot- 
pounds of  work  are  both  the  product  of  force  and  linear  space,  from 
which  it  would  appear  that  these  two  functions  are  substantially  alike, 
but  they  are  of  entirely  different  nature. 

Static  momentum  is  force  multiplied  by  the  geometrical  element 
length,  without  regard  to  velocity  and  time ;  in  which  case  the  force 
has  nothing  to  do  with  the  generation  of  that  length. 

Work  is  force  multiplied  by  the  physical  function  space,  which  is 
generated  by  the  two  elements  velocity  and  time. 

"Work  done  is  expressed  by  the  following  terms  : 

Hauled,  dragged,  raised,  heaved,  cultivated,  tilted,  broken,  crushed, 
thrown,  wrought,  fermented,  labored,  embroidered,  etc.,  or  any  expres- 
sion which  implies  the  three  simple  elements,  force,  velocity  and  time. 

Power  is  the  differential  of  work. 

Work  is  the  integral  of  power. 

The  following  formulas  show  the  different  combinations  of  the  dy- 
namic elements  and  functions.  Should  either  or  both  the  force  and 
the  velocity  be  variable  or  irregular,  the  mean  action  in  the  time  T 
must  be  inserted,  and  the  formulas  will  answer  for  any  kind  of 
operation. 


70 


ELEMENTS  OF  MECHANICS. 


§11-     DYNAMICAL   FORMULAS. 
Force  or  Pressure  in  Pounds. 

p 

w 

F=—  . 

.    1 

JPI=—  .. 

.    3 

v' 

# 

550  IP 

^ 

V 

FT' 

: 

Velocity  in  Feet  per  Second. 

550  H> 

7 

F= 

" 

^     '    '        ' 

•           I 

v  p 

.    6 

F'      -"• 

.     8 

v  — 

—          ... 

FT 

Time  of  Action  in  Seconds. 

rr-       O 

.        .9 

F  8 

.  11 

T  v    • 

550  IP' 

T=™.      . 

.  10 

T_   % 

.  12 

P 

F  V' 

Power  in  Effects. 

P=FV.    . 

.  13 

P=550IP.     . 

.  15 

p_F8 

.  14 

P-K 

.  16 

T  ' 

T'  ' 

Space 

Passed  Through  in  the  Time  T. 

fa  —  v  T 

17 

a    550  yiP 

1Q 

O  —    V    JL.      . 

pr 

•            ..  Xf 

18 

.      -L  (7 

20 

J*  • 

.            .    lo 

s  p.. 

•       i-iW 

Horse  -Power. 

H>      P 

21 

FS 

550' 

550  T'     ' 

.  23 

rp      -^  ^ 

00 

IP  — 

"  550' 

•                 •     —  -• 

550  7* 

' 

Work  in  Foot-Pounds. 

K-FVT. 

-  25            K=F8. 

.  27 

K=  P  T.    . 

.  26 

JT=550IP  7! 

.  28 

EXAMPLES.  71 


It  will  be  observed  in  the  preceding  formulas  that  an  element  is 
never  divided  by  an  element,  but  a  function  is  divided  by  an  element 
only  when  that  function  contains  the  element  divided  with. 

Power  divided  by  velocity  gives  force,  because  power  contains  the 
elements  force  and  velocity  ;  but  power  cannot  be  divided  -by  time, 
because  time  is  not  a  constituent  element  of  power. 

Work  can  be  divided  by  either  one  or  two  of  its  three  constituent 
elements.  When  work  is  divided  by  either  two  of  its  elements,  the 
product  will  be  the  third  element. 

Different  elements  or  functions  cannot  be  added  to  or  subtracted 
from  one  another.  Power  or  space  cannot  be  added  to  or  subtracted 
from  work.  Force,  velocity  or  time  cannot  be  added  to  or  sub- 
tracted from  space. 

When  a  formula  contains  several  terms,  all  the  terms  must  be  of 
the  same  kind  ;  for  instance  : 

Work  K=  T( 

The  terms  within  the  parentheses  are  all  power,  which  multiplied 
by  time  gives  work. 

Mistakes  in  dynamical  formulas  are  easily  detected  by  the  above 
rules. 

No  element  can  be  converted  into  an  element  of  a  different  kind. 

g  12.     EXAMPLES   CORRESPONDING   WITH   THE   FORMULAS. 

Force  or  Pressure  in  Pounds. 

Example  1.  A  power  .P=6400  effects  is  operating  with  a  velocity 
of  V--=  12  feet  per  second.  Required  the  force  F? 


Example  2.  The  piston  of  a  steam-engine  of  IP  =  24  horses  is  mov- 
ing at  the  rate  of  V  =  8  feet  per  second.     Required  the  force  F? 

„    550  IP      550x24      __„ 
F=  —      —  =  -  =  1650  pounds. 
V  8 

Example  S.  A  work  of  K=  3266  foot-pounds  is  accomplished  in  a 
space  #=16  feet.     Required  the  force  F? 


=          =  204  pounds. 


72  ELEMENTS  OF  MECHANICS. 

Example  4-  A  work  of  _fiT=  183600  foot-pounds  was  accomplished 
with  a  velocity  "F"=18  feet  per  second  in  a  time  of  3  minutes,  or 
T=  3  x  60  =  180  seconds.  Required  the  force  F? 

K        183600 


Velocity  in  Feet  per  Second. 

Example  5.  A  body  moves  through  a  space  of  8=  160  feet  in  a 
time  of  T=>  40  seconds.  Required  the  velocity  V? 

^     S     160 

V  =  —  =  -  =  4  ieet  per  second. 
T      40 

Example  6.  A  power  of  P  =  42G6  effects  is  operating  with  a  force 
F=  760  pounds.  Required  the  velocity  F? 

P     4266 

V=  —  =  --  -  =  5.6  feet  per  second. 
F      760 

Example  7.  The  cylinder  of  a  steam-engine  of  IP  =  160  horse- 
power is  24  inches  in  diameter,  and  the  effective  steam-pressure  is  30 
pounds  to  the  square  inch.  Required  the  velocity  of  the  steam- 
piston  ? 

The  area  of  the  piston  is  452.39  square  inches,  which  multiplied  by 
30  pounds  to  the  square  inch  will  be  a  force  of 

F=  13570.8  pounds. 
„    550  IP    550x160 

-^-  =  l35m8-  =  6-5feetperSeC°nd- 

Example  8.  A  work  of  K=  864360  foot-pounds  is  accomplished 
with  a  force  of  F=  68  pounds  in  a  time  of  5  minutes.  Required  the 
velocity  V? 

The  time  T=  5  x  60  =  300  seconds. 

T.     K      864360 
=        ="  =  42-4fe 


Time  of  Action  in  Seconds. 

Example  9.  A  space  of  #=2896  feet  is  generated  with  a  velocity 
of  F=  25  feet  per  second.     Required  the  time  T? 


115.84  seconds. 
25 


EXAMPLES.  73 


Example  10.  A  force  of  .F=  4596  pounds  is  working  through  a 
space  8=  960  feet.  What  time  is  required  for  the  force  to  generate  a 
power  of  P  =  840680  effects  ? 

,    FS    4596x960     . 
T=  -  =  —          —  -  5.25  seconds. 
P        840680 

Example  11  a.  The  stroke  of  a  steam-piston  is  four  feet,  and  the  ef- 
fective pressure  of  steam  is  F=  46360  pounds.  The  power  of  the 
engine  is  EP  =  500  horses.  What  time  is  required  of  the  engine  to 
make  64  double  strokes  ? 

The  space  #=  4  x  2  x  64  =  512  feet. 

F8        46360x512 
r"550g"    550x500 

Example  11  b.  What  time  is  required  to  raise  a  weight  of  200  tons 
to  a  height  of  $=50  feet  with  an  engine  of  IP  =  8  horse-power? 

F=  200  x  2240  =  448000  pounds. 
_,      FS       448000x50 


or  8  minutes  and  29  seconds. 

Example  12.  What  time  is  required  to  accomplish  a  work  of 
JT=  96236000  foot-pounds,  with  a  force  .P=88  pounds,  moving  with 
a  velocity  of  y=1.5  feet  per  second? 

-      96236000 


88x1.5 

or  202  hours  31  minutes  and  6  seconds. 

Assuming  a  workmanday  to  be  1,440,000  foot-pounds,  it  would  re- 
quire about  67  such  units  to  accomplish  the  work ;  that  is  to  say,  one 
man  could  do  the  work  in  67  days,  or  67  men  could  accomplish  it  in 
one  day. 

Power   in  Effects   or  Foot-Pounds. 

Example  13.     A  weight  of  five  tons  is  raised  vertically  at  the  rate 
of  1-j  inches  per  second.     Required  the  power  P? 
The  force  F=  5  x  2240  =- 11200  pounds. 
Velocity  V=  0.125  feet  per  second. 

P=  11200  x  0.125  =  1400  effects. 


74  ELEMENTS  OF  MECHANICS. 

One  man-power  is  50  effects,  and  it  would  require  1400  :  50  =  28 
men  to  raise  five  tons  with  a  velocity  of  1-J-  inches  per  second  at  con- 
tinued work. 

One  horse-power  is  550  effects,  and  it  would  require  1400  :  550 
=  2.55  horse-power  for  the  same  work. 

Example  14-  What  power  is  required  to  lift  a  weight  of  three 
tons  a  space  of  8=  5  feet  in  a  time  of  10  minutes  ? 

.    FS    3x2240x5    , 

F=-f--^W-  56effeots' 

Example  15.     How  many  effects  are  there  in  IP  =  30  horse-power? 
P  =  550  x  30  =  16500  effects. 

Exampk  16.  What  power  is  required  to  do  a  work  of  K=  186000 
foot-pounds  in  one  minute  ?  T=  60. 

p=  186000  =  31000effects 

Space  Passed  Through  in  the  Time  T, 

Example  17.  A  body  moving  with  a  velocity  of  V=  960  feet  per 
second  for  a  time  of  T  =•  5  seconds.  Required  the  space  passed 
through  ? 

#=  T7T=4800  feet. 

Example  IS.  A  power  of  P=  6500  effects  is  operating  for  a  time 
of  T=  12  seconds  with  a  force  F=  240  pounds.  Required  the  space 
passed  through  ? 


Example  19.  To  what  height  can  a  steam-engine  of  IP  =  6  horse- 
power lift  a  weight  of  25  tons  in  a  time  of  5  minutes  ? 

F=  25  x  2240  -  56000  pounds. 
T=  5x60  =  300  seconds. 

,    550xIPr    550x8x300     0_  .  ,    . 
The  hexght  5-  —  jH-  =       56ooo      =  23'6  feet 

Example  20.  A  work  of  K=  7280  foot-pounds  is  to  be  accom- 
plished by  a  force  of  F=  24  pounds.  In  what  space  can  the  force  do 
the  work  ? 


EXAMPLES.  75 


Horse-Power. 

Example  21.      How  many  horse-power  are  there  in  P=  56680 
effects? 


Example  22.     A  weight  of  three  tons  is  to  be  raised  with  a  velocity 
of  F=  6  feet  per  second.     Required  the  horse-power  ? 


Example  23.  A  steam-crane  is  to  be  constructed  to  lift  30  tons  12 
feet  high  in  5  minutes.     Require  1  the  horse-power  ? 
Force  F=>  30  x  2240  =  67200  pounds. 
Time  T=  5  x  60  -  300  seconds. 

F8      67200x12 


55  55300 

Example  24.  What  horse-power  is  required  to  accomplish  a  work 
of  K=  346000  foot-pounds  in  T=5  seconds? 

K       346000 


WORK  IN  FOOT-POUNDS. 

Example  25.  How  much  work  is  accomplished  with  a  force  of 
F=  280  pounds,  moving  with  a  velocity  of  V  =  9  feet  per  second  for 
a  time  of  T=  1200  seconds,  or  20  minutes? 

K=  F  V  T=  280  x  9  x  1200  =  3024000  foot-pounds, 

or  2.1  workmandays. 

Example  26.  How  much  work  can  be  accomplished  by  a  power  of 
P=36  effects  during  T=4  seconds? 

K=  p  T=  36  x  4  =  144  foot-pounds. 

Example  27.  A  weight  of  25  tons  is  lifted  #-18  feet.  Required 
the  work  9 

K=  FS=25x  2240  x  18  =  1008000  foot-pounds. 

Example  28.  How  much  work  is  accomplished  per  minute  by  an 
engine  of  IP  =  48  horse-power? 

K=  550  IP  T=  550  x  48  x  60  -  1584000  foot-pounds. 


76 


ELEMENTS  OF  MECHANICS. 


I  13.     CIRCULAR   OR   ROTARY   MOTION. 

In  this  case  it  is  supposed  that  the  force  F  is  applied  in  the  direc- 
tion of  a  tangent  to  the  circle  of  radius  r  in  feet,  like  that  of  a  belt  or 
rope  over  a  pulley,  or  in  all  kinds  of  gearing. 

n  =  revolutions  of  the  circle  per  minute. 

JV=  total  revolutions  in  the  time  T,  or  for  generating  a  definite  cir- 
•cular  space  S,  and  also  for  the  accomplishment  of  a  definite  work  K. 


DYNAMICAL   FORMULAS   FOR   CIRCULAR   MOTION. 

Velocity  in  the  Periphery  of 
the  Circle  in  Feet  per 
Second. 


V= 


60 


=  0.  10472  r  n. 


.  29 


.  30 


•Revolutions  of  the  Circle  per 
Minute. 


60  F 


„  SI 


Total  Revolutions  JV. 

8 

2  r  r 

K 


.  33 


.  34 


Force  in  Pounds,  acting  in 
the  Periphery. 


5250  IP 


K 


.  35 


.  36 


Horse-Power,  acting  in  the 
Periphery. 

fr*  .  37 


550x60       5250 

F2nrN_  Fr  N 

550  T   =87.5  T' 


.  38 


Work  in  Foot-Pounds,  ac- 
complished in  y  Revolu- 
tions, or  in  the  Time  T. 

.  39 


K- 


60  T 


.  40 


EXAMPLES    FOR   CIRCULAR   MOTION   CORRESPONDING   WITH    THE 
FORMULAS. 

Example  29.  The  radius  of  a  wheel  or  a  crank-pin  is  r  =  2.5  feet, 
and  makes  n  =  56  revolutions  per  minute.  Required  the  velocity  in 
the  circumference? 


60 


- 0.1472  rn  =  0.1472 x 2.5 x 56  =  20.6  feet  per  second. 


STEAM-ENGINES. 


Example  80.  Required  the  velocity  in  the  periphery  of  a  fly-wheel 
of  radius  r  =  8  feet,  and  making  N=  125  revolutions  in  T-  164  seconds? 

,.    2*rN    6.28x8x125 

F=  — — -  =  -  —  =  39.25  feet  per  second. 


§14.     DYNAMICS   OF   STEAM-ENGINES. 

The  following  formulas  are  for  a  double-acting  steam-engine,  of 
which  the  stroke  of  piston  =  s  in  feet. 

F=  force  or  pressure  of  steam  on  the  piston. 

If  the  steam  is  expanded  in  the  cylinder,  F  means  the  mean 
pressure  on  the  whole  piston  throughout  the  stroke  s. 

n  =  double-strokes  per  minute. 

IP  =  horse-power  of  the  engine. 

Velocity  of  the  piston  in  feet  per  second,      V= = — .         .41 

60       30 

__.       2  Fs  n         Fs  n 
Horse-power  of  the  engme,     &-——,——.      .        .42 

^    ZFsnT    FsnT 

Work  done  in  the  time  T,     K= —    — .         .         .43 

60  30 

Work  done  in  ^double-strokes,     K=2Fs&        ,    -  ^        .44 

Let  A  denote  the  area  of  the  steam-piston  in  square  inches,  and 
p  =  mean  steam-pressure  in  pounds  per  square  inch.  Then  the  force 
on  the  piston  will  be  F=  A  p 45 

And  the  horse-power,     EP  =  — — — .  46 

16500 

This  is  the  gross  horse-power  of  the  engine,  including  that  ex- 
pended in  friction  and  working  the  pumps,  and  is  generally  called 

Indicated  Horse-Power. 

The  indicated  horse-power  is  calculated  from  the  indicator  diagram 
or  card  taken  for  that  purpose. 


78 


ELEMENTS  OF  MECHANICS. 


15. 


Load  of  Burden  that  can  be  Carried  by  Man  and 
Animals. 


Carriers. 

Eoad, 
kind  of. 

Force. 
Burden, 
in  pounds. 

Velocity. 
Feet  per 
second. 

Time. 
Hours  per 
day. 

Space, 
in  miles. 

Man 

Good  level.... 

100 

3 

7 

14.3 

Man 

Ordinary     .  . 

95 

2.5 

7 

12 

Man 

50 

35 

10 

238 

Llama  of  Peru 
Donkey  

Mountainous. 
Good  level.... 

100 

300 

3.5 
3.5 

10 
10 

23.8 
23.8 

Donkey  

Mountainous. 

200 

3.5 

10 

23.8 

Mule  

Good  level 

500 

5.0 

10 

34 

Mule.:.  
Horse  

Mountainous. 
Good  level 

400 
300 

4.5 
6 

10 
g 

40.6 
327 

Horse  

300 

45 

g 

245 

Camel  

Deserts 

1000 

3  to  4 

12 

30  to  40 

Elephant  

Ordinary  

1800 

3  to  4 

10 

35 

Man  or  Animal  "Working  a  Machine. 


Working- 
man  or 
animal. 

Machine  which  is 
worked. 

Force, 
in 
pounds. 

Elements. 

Velocity. 
Feet  per 
second. 

Time. 
Hours 
per  day. 

Space, 
in 

feet. 

Functlo 

Power, 
in 

eflect. 

ns. 

Work,  in 
foot- 
pounds. 

F 

V 

T 

s 

P 

K 

Man  

Rope  and  pulley 

50 

0.8 

6 

17280 

40 

864800 

Man  

Crank  

20 

2.5 

8 

72000 

50 

1440000 

Man  

Tread-wheel*  

144 

0.5 

8 

14400 

72 

2073600 

Man  

Tread-  wheel  t  

30 

2.5 

8 

72000 

75 

2160000 

Man  

Draws  or  pushes. 

30 

2 

8 

57600 

60 

1728000 

Horse.... 

Horse-mill  

106 

3 

8 

64800 

318 

6768800 

Horse.... 

Horse-mill  

72 

9 

5 

162000 

648 

11664000 

Horse.... 

4-wheel  carriage. 

154 

3 

10 

108000 

462 

16632000 

Horse..  / 

Revolving         •. 

100 

3 

8 

86400 

300 

8640000 

Mule  J 

mill 

66 

3 

8 

64800 

198 

4276800 

Ass  I 

platform.    * 

33 

3 

8 

64800 

99 

2138400 

*  Axis  horizontal. 


t  Axis  24°  from  vertical. 


HORSE-PO  WER.  79 


I  16.     HORSE-POWER  REQUIRED  TO  DRIVE    DIFFERENT  MACHINES. 
WATER-WORKS. 

For  every  100  gallons  of  water  pumped  per  minute  to  a  vertical  IP 

height  of  100  feet,  requires 5 

For  every  million  (1,000,000)  gallons  pumped  per  24  hours 

to  a  height  of  100  feet,  requires 35 

ROLLING-MILLS.  IP 

For  every  square  foot  of  heated  iron  plate  passing  through 
the  rollers,  requires  .........  5 

Bar-iron  mills.  Two  pairs  of  rough  and  two  pairs  of  fin- 
ishing rollers,  six  puddle  furnaces,  two  welding  furnaces,  making 
10  tons  of  bar  iron  per  24  hours,  rollers  making  70  revolu- 
tions per  minute,  require ,  .  .  .80 

SAW-MILL—ALTERNATIVE. 

For  every  100  square  feet  sawed  per  hour  in  dry  oak,  requires      5 
Dry  pine,  per  100  square  feet,  requires 3 

CIRCULAR-SAWS. 

A  saw  3  feet  in  diameter,  making  300  revolutions  per  minute, 
will  saw  50  square  feet  per  hour  in  dry  oak,  and  requires        .       2 
Dry  spruce,  90  square  feet  per  hour         ...         .         .         .2 

THRESHING-MACHINES. 

Velocity  of  feed-rollers  at  the  circumference,  0.55  feet  per 
second.  Diameter  of  threshing-cylinder,  3.5  feet,  and  4.5  feet 
long,  making  300  revolutions  per  hour,  can  thresh  30  to  40 
bushels  of  oats,  and  from  25  to  35  bushels  of  wheat  per  hour  .  4 

FLOUR-MILLS. 

One  pair  of  mill-stones  4  feet  in  diameter,  making  130  revo- 
lutions per  minute,  can  grind  6  bushels  of  wheat  to  fine  flour 
per  hour    ...........       5 

Can  grind  6  bushels  of  rye  to  coarse  flour  per  hour         .         .       3 
For  every  100  pounds  of  fine  flour  ground  per  hour,  requires  .       1 


80 


ELEMENTS  OF  MECHANICS. 


Fig.  61. 


§  17.     DREDGING-MACHINES. 

The  accompanying  illustrations  of  dipper  and  grapple  dredges 
are  furnished  by  the  American  Dredging  Company  of  Philadelphia. 

Dipper-Dredge. — Fig.  61  represents  the  or- 
dinary dipper-dredge,  consisting  of  one  scoop, 
worked  with  a  triple  cnain  wound  on  a  15-inch 
drum,  and  driven  by  a  pair  of  engines  10  inches 
in  diameter  by  15  inches  stroke  of  cylinders. 
Under  ordinary  work  the  scoop  makes  30  to  40 
dips  per  hour,  and  takes  up  about  two  cubic 
yards,  or  three  tons,  of  materials  each  dip. 

The  dipper-dredge  is  used  in  harbors  and  docks,  and  also  in  rail- 
road excavations. 

Fig.  62.  Grapple-Dredge. — Fig.   62   represents    the 

grapple-dredge,  consisting  of  a  double  scoop 
opening  in  the  bottom  like  a  mouth,  and  takes 
up  about  five  tons  of  materials  each  grapple.  It 
is  worked  by  a  single  chain  wound  on  a  drum 
three  feet  in  diameter,  with  a  pair  of  engines 
MTEKT  iMPHovEDTBRAppiE-oREDaE.-  -^  ^c^eg  diameter  by  20  inches  stroke  of  cylin- 
ders. Under  ordinary  work  it  makes  50  to  60  grapples  per  hour. 

Ladder-Dredge. — The  ladder-dredge  consists  of  an  endless  chain 
upon  which  a  number  of  buckets  are  fixed  and  work  continually  like 
a  Noria.  This  appears  to  be  the  best  form  of  dredge  for  deepening 
harbors,  but  is  not  so  well  suited  for  docks,  where  the  dipper  and 
grapple  dredges  are  the  best. 


FORMULAS   FOR   THE    LADDER-DREDGE. 


700  IP 


F- 


550  Tk 


IP  =  horse-power  required  for  excavating  and  raising  the  materials. 

T  =tons  of  materials  excavated  and  raised  per  hour. 

h  =  height  in  feet  to  which  the  excavated  materials  are  raised. 

F=  force  in  pounds  required  to  feed  the  dredge  ahead. 

V  =  velocity  of  the  buckets  in  feet  per  second. 

k  =  0.1  for  hard  clay  with  gravel.  |  k  =  0.05  for  common  clay  and  sand. 
k  =  0.07  for  hard  pure  clay.  j  k  =  0.04  for  soft  clay  and  loose  sand. 


STEAMSHIP  PERFORMANCE. 


81 


\  18.    POWER  REQUIRED  TO  PROPEL  STEAMBOATS  AT  DIFFERENT 
SPEEDS. 

M=  nautical  miles  or  knots  per  hour. 

T  =  displacement  in  tons,  which   must  be    well  proportioned  for 


IP  =  horse-power  required  to  propel  the  vessel  M  miles  per  hour. 

w~^f- 


Tons. 

2 

4 

NAU 

6 

TICAL  ] 

8 

MCILES  C 

1O 

>R  KN0 

12 

rs  PER 
14 

HOUB. 

16 

18 

2O 

T 

IP 

IP 

IP 

IP 

H> 

IP 

H> 

IP 

IP 

IP 

1 

0.035 

0.280 

0.794 

2.240 

4.386 

7.792 

12.03 

17.92 

25.60 

35.09 

2 

0.055 

0.444 

1.260 

3.555 

6.960 

10.08 

19.09 

28.44     40.63 

54.68 

3 

0.075 

0.598 

1.651 

4.787 

9.123 

13.21 

25.00 

38.30     53.25 

72.98 

4 

0.084!  0.673 

1.900 

2.389 

11.05 

15.20 

30.31 

43.11 

64.51 

88.41 

5 

0.102 

0.818 

2.207 

6.550 

12.22 

17.65 

35.17 

52.40 

74.85 

97.76 

6 

0.115 

0.924 

2.620 

7.392 

14.47 

20.96 

39.70 

59.13 

84.48 

115.8 

7 

0.128 

1.025 

2.906 

8.198 

16.05 

23.25 

44.03 

65.50 

93.70 

128.4 

8 

0.130 

1.120 

3.176 

8.96 

17.54 

31.17 

48.12 

68.70 

102.4 

140.3 

9 

0.151  11.211 

3.430 

9.690 

19.00 

27.44 

52.10 

77.52 

110.1 

152.0 

10 

0.162 

1.300 

3.684 

10.40 

20.35 

29.47 

55.82 

83.20 

118.8 

162.8 

15 

0.213 

1.702 

4.827 

13.62 

26.66 

38.62 

73.50 

108.9 

156.0 

213.2 

20 

0.258 

2.064 

5.845 

1651 

32.25 

46.76 

88.88 

132.1 

188.5 

258.0 

30 

0.338 

2.704 

7.675 

21.63 

42.42 

60.40 

116.5 

173.0 

248.0 

339.4 

40 

0.409 

3.272 

9.280 

26.17 

51.25 

74.24     140.5 

209.3 

300.0 

410.0 

50 

0.474 

3.792 

10.79 

30.33 

59.53 

86.32     163.5 

242.6 

346.2 

476.2 

60 

0.538 

4.304 

12.20 

34.43 

67.35 

97.60 

185.0 

285.4 

393.5 

538.8 

70 

0.597  4.676 

13.55 

38.21 

74.85 

108.4 

205.5 

305.7 

437.0 

598.8 

80 

0.650|  5.200 

14.8 

41.60 

81.60 

118.4     224.0 

332.8 

476.0 

652.8 

90 

0.705 

5.640  16.00 

45.22 

88.40 

128.0     242.5 

361.7 

516.0 

707.2 

100 

0.755 

6.040   19.40 

48.4 

94.5 

163 

259 

387 

551 

756 

150 

0.990 

7.720  22.51 

61.76 

124.0 

180.0 

341.5 

494.1 

724.0 

992.0 

200 

1.200 

9.600     32.5 

76.9 

150 

260 

412 

615 

875 

1201  i 

300 

1.575 

10.60     42.4 

100 

196 

340 

540 

806 

1146 

15731 

400 

1.910 

15.28     51.4 

122 

238 

412 

654 

976 

1402 

1907 

500 

2.213 

17.70     59.6 

141 

276 

478 

759 

1131 

1611 

2213 

600 

2.50 

20.00     67.2 

160 

313 

540 

856 

1280 

1820 

2500 

!   700 

2.780 

22.24     74.6 

177 

377 

599 

938 

1417 

•    2016 

2770 

800 

3.025 

24.20     81.5 

194 

388 

654 

1038 

1548 

2206 

3026 

1000 

3.500 

28.00 

94.6 

225 

439 

759 

1206 

1798 

2560 

3514 

1500 

4.6111  36.88 

124 

295 

575 

995 

1580 

2355 

3352 

4605 

2000 

5.600 

44.80 

150|       356 

696 

1205 

1913 

2854 

4060 

5570 

3000 

7.350 

58.80 

197        467 

913 

1582 

2508 

3740 

5318 

7300 

4000 

8.87 

70.96 

238 

567 

1105 

1912 

3038 

4530 

6444 

8847 

6000    11.441  91.52 

303|       742 

1448 

2507 

3981 

5935 

8446 

11586 

82  ELEMENTS  OF  MECHANICS. 


§19.    FRICTION. 

Sliding-  Friction  is  the  force  required  to  rub  or  slide  one  sur- 
face upon  another.  For  the  same  kind  of  surfaces  the  force  of  fric- 
tion is  proportionate  to  the  pressure  of  contact,  and  independent  of 
the  velocity  with  which  the  rubbing  or  sliding  body  moves.  Within 
certain  limits  the  friction  is  also  independent  of  the  extent  of  surface 
in  contact. 

Fig-63-  Let  a  b  represent  a  horizontal  surface 

_         on  which  is  placed  a  body  W  in  close  con- 
_        */y^      tact  with  a  b.     The  body  W  is  attached  to 
~\^/l      a  weight  /  by  a  rope  over  a  pulley.     Ad- 
fps    just  the  weight  /  so  that  it  will  barely 
K   '    move  the  body  W;  then  /  is  the  force  of 
friction  of  the  surfaces  in  contact.     If  the 

body  W  be  started  with  a  certain  velocity  V,  the  weight  /  will  con- 
tinue that  velocity  uniformly  ;  but  if  it  is  greater  than  the  friction, 
the  velocity  will  be  accelerated. 

TP=  weight  of  the  body  in  pounds,  which  is  the  force  or  pressure 

of  contact  of  the  friction  surfaces. 
f  =  force  in  pounds  of  the  friction. 
#  =  ratio  of  W  and/,  or  the  coefficient  of  friction. 
V=  velocity  in  feet  per  second  of  the  motion. 
S=  space  of  motion  in  feet. 


=  =  n=. 

W  & 

Friction  power,  P=f  V. 

Friction  horse-power,       IP  =  -  —  . 
550 

Work  of  friction,  K=f  8. 

The  work  expended  on  friction  is  generally  converted  into  heat, 
which  is  not  utilized,  but  lost.  It  is  therefore  of  great  importance  in 
the  working  of  machinery  to  reduce  the  work  of  friction  to  the  lowest 
possible  amount,  for  which  reason  lubricating  substances  are  intro- 
duced between  the  friction  surfaces,  such  as  powdered  graphite  or 
soapstone,  and  all  kinds  of  fatty  substances,  such  as  oil,  tallow,  lard, 
soap,  etc.  ;  all  of  which  reduce  the  friction  coefficient,  but  to  a  dif- 
ferent degree,  depending  on  how  and  on  what  kind  of  surfaces  the 
lubrication  is  used. 


FRICTION.  83 


The  force  of  friction  can  be  ascertained  only  by  experiments  which 
have  been  made  by  Coulomb,  Vince,  G.  Rennie,  N.  Wood  ;  and  the 
most  complete  and  reliable  experiments  on  friction  were  made  by 
Arthur  Morin  in  the  years  1831,  '32,  and  '33,  at  the  expense  of  the 
French  government. 

Friction  surfaces  must  be  either  plane,  cylindrical,  spherical,  con- 
ical or  any  figure  concentric  with  an  axis  of  rotation. 

£20.     PLANE    FRICTION   SURFACE. 

The  friction  coefficient  for  plane  surfaces  is  best  determined  by  a 
body  sliding  on  an  inclined  plane. 

h  =  height  of  the  inclined  plane.  Fig  M_ 

I  =  length,  and  b  =  base. 
W=  weight  of  the  sliding  body. 
f=  force  of  friction,  which  is  equal  to  the  force 
of  gravity  acting  to  draw  the  body  down 
on  the  inclined  plane. 

The  inclined  plane  is  so  elevated  or  adjusted  that  the  body  barely 
moves  by  its  own  weight. 

Let  x  denote  the  angle  of  the  inclined  plane  with  the  horizon,  then 
the  friction  coefficient  will  be 


From  the  law  of  statics  we  have 

W:f=l:h,    and     Wk=jl. 


Friction  coefficient  <P  =  sin.x  -  -  =  -^—  . 

The  friction  coefficient  is  independent  of  the  extent  of  areas  in  con- 
tact until  near  the  point  of  abrasion. 

When  the  Great  Eastern  was  to  be  launched  it  was  found  that  she 
could  not  be  pulled  down  on  her  ways  with  the  greatest  force  that 
could  be  applied  ;  whereupon  Mr.  Brunell  invited  George  Stephenson 
to  come  to  the  launch  and  give  advice  on  the  same.  Upon  his  arrival, 
George  Stephenson  remarked  that  the  weight  of  the  ship  had  exceeded 
the  abrasion  on  the  ways,  and  that  the  ship  could  not  be  launched 
without  increasing  the  area  of  the  ways  ;  which  was  accordingly  done, 
and  the  ship  went  off  with  considerable  ease. 


ELEMENTS  OF  MECHANICS. 


221.    CYLINDRICAL   FRICTION   SURFACE. 

Axles  and  shafts,  bearing  in  journals,  are  generally  cylindrical. 


Fig.  65. 


\ 


Rw 


pressure    in    the 


W=  weight    or 
journal. 

w  =  weight  or  force  applied  to  give 
rotation  or  motion. 

R  =  radius  upon  which  the  force  w 
acts,  in  feet. 

r  =  radius  in  feet  of  the  journal, 
upon  which  the  force  of  fric- 
tion/ acts. 

n  =  number    of    revolutions    per 

minute. 
R  :  r  =/ :  w    and    R  w  =/r. 

,    Rw  =/r 

T  R 


Friction  coefficient   $  =  — 


Wr 


Horse-power  of  friction    IP  = 


33000 


33000 


rnf     rn  W$ 
5252      5252  ' 


\  22.     CONICAL  FRICTION    SURFACE. 

This  illustration  represents  a  vertical  shaft  supported  on  a  conical 
pivot. 

W=  force  or  pressure  in  pounds  in  the  direc- 
tion of  the  centre  line  of  the  shaft. 
r  =  radius  of  the  base  of  the  cone  in  feet. 
x  =  half  the  angle  of  the  cone. 
f  =  force  in  pounds  of  the  friction  acting  on 

the  mean  radius,  0.707  r. 
w  =  pressure  in  pounds  on  the  conical  friction 

surface. 
n  **  number  of  revolutions  per  minute. 

W   -=   W  COS6C.X. 

/-**. 


FRICTION.  85 


In  this  case  the  friction  coefficient  is  the  same  as  for  cylindrical 
journals. 

0.70711  rnw  $     rnw$ 
Friction  horse-power  I 


For  a  flat  circular  surface  x  =  90°  and  cosec.x  -  1. 

Friction  horse-power  IP  =  -  -. 

7427.5 

|  23.     HORIZONTAL  SHAFT. 

Fig.  67. 

When  the  shaft  is  horizontal  and  the  pres- 
sure vertical,  the  pressure  on  the  conical  fric- 
tion surface  will  be  w  =  W  sec.x. 


7427.5 


|  24.    SPHERICAL  FRICTION  SURFACE. 

For  a  spherical  pivot  the  mean  radius  is  at  half  the  height  of  the 
segment. 

R  =  radius  of  the  sphere. 
r  =  mean  radius  of  friction. 
W=  pressure  on  the  friction  surface. 

rnW* 


The  friction  coefficient  in  this  case  will  be  the  same  as  that  for 
cylindrical  journals. 


ELEMENTS  OF  MECHANICS. 


TABLE  I. 

§25.     SUMMARY   OF   MORIN'S   EXPERIMENTS   ON    FRICTION. 

=  fibres  moving  parallel  on  one  another. 

+  fibres  moving  at  right  angle  to  fibres. 

±  end  of  fibres  moving  parallel  to  fibres. 

-i-  end  of  fibres  moving  at  right  angle  to  fibres. 

I  fibres  moving  end  to  end. 

Xaterimls. 

Condit'n 

In  motion. 

Martin--. 

Moving. 

Stationary. 

of  fibres. 

Lubrication 

Coef.  *. 

Angle  z. 

Coefc  *. 

Angle  x. 

Oak 

Oak  

= 

non 

0.48 

25°  39' 

0.62 

31°  48' 

Oak  

O.ik  

= 

Dry  soap 

0.16 

9      6 

0.44 

23    45 

Oak 

Oak  

I 

non 

0.34 

18    47 

0.54 

28    22 

Oak 

Oak 

I 

Water 

0.25 

14      3 

0.71 

35    23 

Oak  

Oak  

non 

0.19 

10    46 

0.43 

23    16 

Elm  

Oak  

= 

non 

0.43 

23    17 

0.38 

20    49 

Elm 

Oak  

I 

non 

0.45 

24    14 

0.57 

29    41 

Elm 

Oak  

non 

0.25 

14      3 

Ash,  fir  

Oak  

=    ' 

non 

0.38 

20    48 

0.53 

27    56 

Iron 

Oak  ..     .. 

_ 

non 

0.62 

31    48 

0.62 

31    48 

Iron 

Oak    . 

Water 

0.26 

14    35 

0.65 

33       9. 

Iron 

Oak 

= 

Dry  soap 

0.21 

11    52 

• 

Cast-iron... 

Oak. 



Water 

0.22 

12    25 

0.65      j  33      2 

Cast-iron... 

Oak  

= 

Dry  soap 

0.19 

10    46 

CoDDer 

Oak  

_ 

non 

0.62 

31    48 

0.62        31     48 

Iron 

Elm  

= 

non 

0.25 

14      3 

Cast-iron... 

Elm  

= 

non 

0.20 

1L    19 

Leather.... 

Oak  

= 

non 

0.27 

15      7 

Leather.... 

]  Cast-  

_L 

non 

0.56 

25    15 

Leather.... 

1  iron  

± 

Water 

0.36 

19    48 

Leather.... 

|  or  

± 

Wat.  oil 

0.23 

12    58 

Leather.... 

J  Brass.... 

± 

Oil 

0.15 

8    32 

Hemp  

Oak  

= 

non 

0.52 

27    29 

0.80 

38    40 

Hemp  

Oak  

I 

Water 

0.33 

18    16 

0.87 

41       2 

Iron  

Iron  

= 

non 

0.44 

23    45 

Iron- 

Brass 

_ 

0.18 

10    13 

Cast-iron... 

Brass  

« 

non 

0.15 

8    32 

Brass  

Brass.... 

« 

non 

0.20 

11     19 

Brass  Cast-iron.. 

H 

non 

0.22 

12    25 

Brass  

Iron  

" 

non 

0.16 

9      6 

/Hard.... 

Hard...., 

"      f 

Well      i 

0.07  to 

4      1 

0.10  to 

5    43 

j  Wood  or  Wood  or  I 

«     I 

greased  J 

0.08 

4    35 

0.15 

8    32 

1  Iron  llron  ' 

" 

slightly 

0.15 

8    32 

FRICTION. 


87 


TABLE  II. 
3  26.    Friction  in  Shaft-journals.—  Morin. 

Materials 

in  Shaft.          !   in  Journal-box. 

Lubricating  substances. 

Coefficient  *  when 
greased  with  oil 
perpet    at  intervals. 

Cast-iron  
Cast-iron  
Cast-iron  
Cast-iron  
Wrought-iron.. 
Wrought-iron.. 
Wrought-iron.. 
Wrought-iron.. 
Brass 

Cast-iron  
Cast-iron  
Cast-iron  
Lignum  vitae  ... 
Cast-iron  
Brass   -  

Coated  w 
Oil  and  v 
Coated  w 
Oil  or  hoj 
Coated  wi 
Coated  wi 
Greasy  ar 
Scarcely 
Coated  w 
With  hog 
Oil  or  tal 
Hog's  lar 

th  grease  
'ater.. 

0.054 
0.28 
0.19 

0.07  to  0.08 
0.08 
0.054 
0.10 
0.07  to  0.08 
0.09 
0.19 
0.25 
0.10 
0.09 
0.048 
0.04 

th  asphaltum.. 

th  grease  

0.054 

Brass- 
Brass.. 

d  wet  

th  oil 

Brass  

Bra«s 

'slard  

Brass  

Cast-iron  
Lignumvitae.... 

ow  
i  



Lignuravitse.... 

\  27.    Friction 

TABLE  III. 
Coefficient  by  Different  Observers. 

Moving. 

Materials. 

1          Stationary. 

Condition  of 
surface. 

Coeffi- 
cient 
*. 

Angle 
'  limit. 

Soft  calcareous 
Hard  calcareou 
Common  brick 
Soft  calcareous 
Freestone  

stone... 
?  stone- 

Soft  cal 
Hard  c 
Commo 
Soft  cal 
Freesto 
Freesto 
Hardc 
Calcare 
Pavemt 
Beaten 
Bed  of 
Bed  of 
Calcare 

.  stone  
al.  stone... 
n  brick.... 
.  stone  
ne  

Well-dressed... 
Well-dressed... 
Common  

0.74 
0.75 
0.67 
0.74 
0.71 
0.66 
0.58 
0.78 
0.58 
0.33 
0.51 
0.34 
0.36 

36°  W 
38    52 
33    50 
36    30 
35    23 
33    26 
30      7 
37    58 
30      7 
18     16 
27       2 
18    47 
32    13 

stone... 

Fresh  mortar... 
Smooth  

ne  

Fresh  mortar... 
Polished  
Rough  
Common  
Common  
Dry  

Hard  calcareous  stone- 
Calcareous  stone  
Wood-box.  

al.  stone... 
ous  stone., 
nts  

earth  
clay  
clay  
ous  stone.. 

Oak 

Drv 

ELEMENTS  OF  MECHANICS. 


TABLE  IV. 

2  28.    Friction  Coefficients  for  Different  Pressures  up  to  the 
Limit  of  Abrasion. 

FBOM  EXPERIMENTS  BY  MR.  G.  REXXIE. 

("  Philosophical  Transactions,"  1829.) 


Pressure  per 
square  inch. 

Wrought-iron  up- 
on wrought-iron. 

Wrought-iron 
upon  cast-iron. 

Steel 
upon  cast-iron. 

Brass 
upon  cast-iron. 

32.5 

0.140 

0.174 

0.166 

0.157 

187 

0.250 

0.275 

0.300 

0.255 

240 

0.271 

0.292 

0.233 

0.219 

277 

0.285 

0.320 

0.340 

0.214 

315 

0.297 

0.329 

0.344 

0.211 

336 

0.312 

0.333 

0.347 

0.215 

373 

0.350 

0.351 

0.351 

0.206 

411 

0.376 

0.353 

0.353 

0.205 

448 

0.395 

0.365 

0.354 

0.208 

485 

0.403 

0.366 

0.356 

0221 

523 

0.409 

0.366 

0.357 

0.223 

560 

Abrasion. 

0.367 

0.358 

0.233 

597 

0.367 

0.359 

0.234 

635 

0.367 

0.367 

0.235 

672 

0.376 

0.403 

0.233 

709 

0.434 

Abrasion. 

0.234 

747 

Abrasion. 

0.235 

784 

0.232 

821 

0.273 

',  29.     ROLLING  FRICTION. 

Rolling-friction  is  the  resistance  of  uneven  surfaces  rolling  on  one 
another,  like  that  of  a  wheel  rolling  on  a  road.  The  coefficient  of 
rolling  friction  represents  the  unevenness  of  the  surfaces  in  contact, 
for  if  these  surfaces  were  perfectly  hard  and  smooth,  there  should  be 
no  rolling-friction. 

Smooth  Wheel  on  Irregular  Hard  Road. 

A  wheel  of  radius  R  and  weight  W 
rolling  on  an  uneven  road,  and  strikes 
a  projection  at  a,  which  presents  a  resist- 
ance/ to  the  forward  motion. 

/=  Wtan.z. 


Fig.  69. 


FRICTION. 


89 


This  resistance  will  be  diminished  to  nothing  when  the  centre  of 
the  wheel  is  vertical  above  a,  after  which  a  similar  force  acts  with  the 
motion  until  the  wheel  strikes  the  second  projection  b.  The  forces  of 
the  projection  thus  act  alternately  against  and  with  the  motion,  so 
that  there  would  be  no  force  lost ;  but  when  the  wheel  strikes  the 
projection,  work  is  performed  in  crushing  or  wearing  the  surfaces  in 
contact ;  and  it  is  the  force  of  this  work  which  makes  the  rolling- 
friction. 

Irregular  Wheel  on  a  Smooth  Hard  Road. 

The  wheel  may  also  be  irregular  and  run  on  a 
smooth  road,  or  both  the  wheel  and  road  may  be 
-irregular.  In  either  case  the  rolling-friction  is 
the  force  of  the  work  expended  in  wearing  the  sur- 
face in  contact. 


Fig.  71. 


On  Soft  or  Muddy  Roads. 

When  a  wagon  is  run  on  a  soft  or  muddy 
road  the  path  of  the  wheels  represents  the  work 
done  by  the  rolling- friction. 

This  rolling-friction  can  be  measured  on  an  in- 
clined muddy  road  by  loading  the  wagon  until  it 
will  barely  move  by  its  own  weight. 
X=  angle  of  inclination  of  the  road. 
W=  weight  of  the  load  and  wagon. 

/=  friction  resistance  in  the  mud,  omitting  the  rotary  friction  in 
the  axis. 

/=  Wsin.x  =  W$. 

Kolling-friction  coefficient  <?  =  -=—=  sin.x. 
W 

Hard  Wheel  on  Elastic  Road. 

A  hard  wheel  running  on  a  perfectly  elastic 
road  will  leave  no  path  behind,  and  there  is 
thus  no  work  performed  by  rolling-friction,  be- 
cause the  forces  of  the  elastic  road  act  equally 
for  and  against  the  motion. 

Elastic  Wheel  on  Hard  Road 

In  this  case  the  wheel  will  be  compressed  on 
the  road,  and  the  forces  acting  with  and  against 
the  motion  will  be  alike,  and  there  will  be  no 
rolling-friction  if  the  wheel  is  perfectly  elastic. 


Fig.  72. 


8* 


90 


ELEMENTS  OF  MECHANICS. 


A  Load  on  Rollers. 


Fig.  74. 


w 


(J    ()    O 


may 


be 


A  load  moved  on  rollers  placed 
loose  on  the  road  will  move  twice  as 
fast  as  the  rollers  ;  and  the  roller  left 
behind  is  placed  in  front  of  the  load 
continually  until  the  rolling  distance 
is  completed.  Any  number  of  rollers 

used  ;  and  sometimes  they  are  connected  by  rods,  like  in  slip 
Fi    ?5  railways  for  hauling  up  ships. 

The  ship  will  move  twice  as 
fast  as  the  roll-carriage.  The 
length  of  the  ship  is  L,  and 
the  distance  to  be  hauled  is 
D;  the  length  I  of  the  roll- 
carriage  must  be 


W=  weight  of  the  ship  and  carriage. 
x  =  angle  of  inclination  of  the  railway. 
F=  hauling  force. 

F  =  W  sin.x.,  omitting  rolling-friction,  which  may  be  about  5  per 
cent,  of  W. 

It  requires  nearly  double  the  force  to  start  the  motion  that  is  to 
continue  the  same. 

On  hard  roads  the  rolling-friction  is  proportionate  to  the  load  and 
inversely  as  the  diameter  of  the  wheels,  but  is  independent  of  the 
width  and  number  of  wheels. 

On  soft  or  muddy  roads  the  rolling-friction  increases  slightly  with 
the  load  and  diminishes  nearly  as  the  width  of  the  wheel. 

§30.     CYLINDER   ROLLING-FRICTION. 

The  rolling- friction  of  smooth  surfaces  in  contact  is  very  small, 
which  circumstance  is  sometimes  utilized  by  running  shafts  on 
cylinders  or  rollers,  as  represented  by  Fig.  76.  The  shaft  a  runs  on 
the  rollers  b,  b. 

TF=  pressure  of  the  shaft  on  the  rollers. 
JR  =  radius  of  the  rollers. 
r  =  radius  of  the  roller-journals. 
r'  =  radius  of  the  shaft. 
0  =  coefficient  of  friction  in  roller-journals. 
x  =  angle  as  shown  on  the  illustration. 
/=  force  of  friction  on  the  shaft  radius  r' . 


Fig.  76. 


FRICTION. 


91 


w$ 

Fig.  77. 

f 

C°^ 

Rsec.x'                                       | 

The  weight  of  the  shaft  a  may  also  be  supported  by 

7&\ 

one  roller  b,  as  represented  by  Fig.  77,  in  which  case 

to) 

/- 


R 


FRICTION-GEAR. 

For  light  work  motion  can  be  transmitted  from  one  shaft 
to  another  by  forcible  contact  of  two  wheels,  Fig.  78. 
W=  pressure  of  contact. 
$  =  coefficient  for  sliding-friction. 
/=  tangential  force  transmitted  on  the  wheels. 


GROOVED   FRICTION-GEAR. 


The  tangential  force  of  friction-gear  can  be  increased  T 
by  grooving  the  peripheries  of  the  wheels,  as  shown  by 
the  illustration. 

x  =  angle  of  the  groove. 


Fig.  79. 


=  W$  cosec.\x. 


\  31.     ROLLING-MILL. 

The  rolling-friction  in  a  rolling-mill  is  the  force  of  the  work  in 
compressing  the  body  passing  between  the  rollers.  By  knowing  the 
motive-power  and  weight  of  the  fly-wheel,  the  rolling  work  and  fric- 
tion can  be  determined  by  their  performance. 

Assume  the  mill  to  be  worked  by  a  steam-engine  attached  direct 
to  the  fly-wheel  shaft  and  roller,  and 
F=  force  of  steam-pressure  on  the  pis-  Fi&- 80- 

ton. 

8=  length  of  stroke. 
W=  weight  of  the  fly-wheel. 
X=  radius  of  gyration  of  fly-wheel. 
n  =  number  of  revolutions  per  minute 
when  the  body  enters  the  rollers. 
n'  =  revolutions  when  the  body  leaves 

the  rollers. 

JV"=  revolutions  in  which  the  body  was 
rolled. 


92  ELEMENTS  OF  MECHANICS. 

The  work  of  the  steam-engine  in  rolling  the  object  will  then  be 

£  =  2  F8N  in  foot-pounds.  .     1 

The  work  done  by  the  fly-wheel  in  the  same  operation  will  be 
W  Y'1 


The  whole  work  in  rolling  the  body  will  then  be 


T=  thickness,  and  1  =  length  of  the  body  when  entering,  and 
t  =  thickness,  and  L  =  length  when  leaving  the  rolls. 
/=  force  of  compression. 
d  =  length  of  the  object  pressed  by  the  rolls. 
The  work  of  compression  will  be 


which  should  be  equal  to  the  work  expended  by  the  engine  and  fly- 
wheel. 


The  work  JTmast  be  calculated  by  Formula  3,  and  inserted  for  K 
in  Formula  5,  for  finding  the  force  of  compression  of  the  rollers. 
All  linear  dimensions  are  expressed  in  feet  and  forces  in  pounds. 

|32.     TRACTION   ON   LEVEL   ROADS. 

Traction  on  roads  is  the  force  required  to  pull  or  move  a  load  on  a 
horizontal  road,  and  which  includes  both  rolling-  and  axle-frictions. 
On  very  smooth  and  hard  roads,  like  that  of  a  train  on  steel-rails  in 
good  order,  the  rolling-friction  is  a  small  item  of  the  axle-friction.  ' 

W=  weight  of  the  load  moved  on  the  road. 
F=  force  of  traction. 
/=  traction  coefficient. 

/--£,        F-Wft  and    TF=y. 

The  following  table  gives  the  traction  coefficient  on  different  kinds 
of  horizontal  roads  : 


TRACTION. 


93 


TABLE  V. 
Traction  Coefficients. 


Carriage*. 

Roads. 

Traction/. 

Railway  trains.  

New  steel  rails  

0.0025 

00030 

Railway  trains.        .         ... 

Worn  iron  rails  

00040 

Smooth  stone  pavement  

0  0048 

Wagons  

Good  street  pavement.  

0.0080 

Wagons 

Turnpikes-  

00120 

Coarse  gravel  «  

0020 

Wagons  

Common  bad  roads.  

0.060 

0200 

Artillery  wagons  

Good  dry  sand  

0.0260 

Wet  sand 

00250 

00195 

£3     Cart  without  springs  

Moist  gravel  
Dry  gravel  

0.0210 
0.0200 

00245 

0  0272 

3    Carriage  with  6  wheels  
H    Two  carriages  with  12  wheels. 

Sand,  rutted  
Muddy  road  

0.0463 
0.0476 

\  33.     TRACTION  ON  INCLINED  ROADS. 

On  inclined  roads  the  weight  of  the  load  acts  against  or  with  the 
traction  as  the  direction  of  motion  is  up  or  down  the  incline. 
x  =  angle  of  inclination  of  the  road  with  the  horizon. 
F=  tractive  force  up  or  down  the  inclined  road. 

F  =  Wfcos.x  ±  Wsin.x. 

F  =  W(fcos.x±sin.x). 

-fwhen  the  motion  is  upward.         -when  the  motion  is  downward! 

The  road  may  be  so  inclined  that  no  tractive  force  is  required  to 
move  the  load  down,  which  will  happen  when 

/  cos.x  <  sin.x. 

The  load  is  in  equilibrium  on  the  road  when  f  cos.x  =  sin.x,  in 
which  case  it  requires  a  force  of 

F  =  W  sin.x  to  draw  the  load  up. 


94  ELEMENTS  OF  MECHANICS. 

When  sin.x.  >f  cos.x  it  requires  a  force  of 

F=(sin.x-fcos.x)W 

to  hold  the  load  or  to  prevent  it  from  moving  down  the  incline  by  its 
own  weight. 

3  34.     ADHESION  ON  HORIZONTAL  RAILS. 

Adhesion  on  rails  is  the  force  of  sliding  friction  of  the  locomotive 
wheels.  This  force  of  adhesion  is  equal  to  the  weight  of  the  locomo- 
tive multiplied  by  the  friction  coefficient  in  Table  I. 

MO  =  weight  of  the  locomotive  in  pounds. 
A  =  force  of  adhesion,         A  =  w  $. 

This  force  must  be  greater  than  the  tractive  force  required  to  move 
the  train,  in  order  to  enable  the  locomotive  to  go  ahead  without 
sliding  the  wheels. 

W=  weight  of  the  train  in  pounds. 
/=  coefficient  of  traction,  Table  V. 

There  must  be  w  $>  Wfio  produce  motion. 

The  locomotive  wheels  will  slide  on  the  rails  when  Wf>  w  $. 

TABLE  VI. 
Coefficient  of  Adhesion  on  Rails. 


Condition  of  Rails. 

0 

0301 

Very  dry 

0224 

Under  ordinary  circumstances  

0.20 

In  wet  weather  

0141 

With  snow  or  frost  

0100 

ADHESION  ON  INCLINED  RAILS. 

The  inclination  of  the  track  diminishes  the  force  of  adhesion  as  the 
cosine  for  the  angle,     x  =  angle  of  inclination  of  the  track. 

Adhesion,  A  =  w  $  cos.x. 
The  force  of  traction  of  the  locomotive  is  limited  to  the  force  of 


RAILROADS.  95 


adhesion,  and  when  a  train  is  running  up  or  down  an  inclined  track 
the  tractive  force  must  be 

F  =  Wfcos.x  i  W  sin.x  =  or  <  w  $  cos.x. 

+  when  the  traction  is  upward. 
—  when  the  traction  is  downward. 

The  traction  and  gravity  of  the  train  is  in  equilibrium  when 

Wf  CQS.X  =  Wsin.x, 
or  when  f  cos.x  =  &in.x. 

The  train  will  run  down  the  track  by  its  own  force  of  gravity, 
when  sin.x>f  cos.x. 

The  resistance  of  wind  to  the  train  is  not  included  in  the  formulas 
for  force  of  traction. 

Experiments  have  shown  that  the  force  of  traction  increases  slightly 
with  the  velocity ;  which  is  mostly  due  to  resistance  of  the  air. 


TRACTION-POWER. 

The  tractive  power  is  equal  to  the  force  of  traction  F  multiplied 
by  the  velocity  in  feet  per  second ;  and  if  desired  in  horse-power 
divide  the  product  by  550. 

When  the  velocity  is  expressed  in  miles  M  per  hour,  the  horse- 
power will  be 


96 


ELEMENTS  OF  MECHANICS. 


§35.    BELT   AND   PULLEYS. 

The  best  and  simplest  mode  of  transmitting  motion  from  one  shaft 
to  another  is  by  a  belt  and  pulleys,  which  is  very  extensively  used  and 
it  gives  the  smoothest  motion.  The  motion  is  transmitted  by  the 
frictional  adhesion  between  the  surfaces  in  contact  of  the  belt  and 
pulleys,  for  which  reason  that  friction  must  be  greater  than  the  ten- 
sion of  the  belt,  otherwise  the  belt  will  slip  and  fail  to  transmit  all 
the  motion  due  from  the  driving  pulley.  There  is  always  some 
slip  in  belt  and  pulleys,  for  which  reason  that  mode  of  transmission  is 
not  positive  or  exact,  and  cannot  be  used  where  precise  motions  are 
required. 

Fig.  81  represents  a  belt  transmitting  motion  between  two  parallel 
shafts  a  and  b.     If  the  motion  is  transmitted  from  a  to  b,  the  pulley 
Fi    „  D  is  called  the  driving  pulley,  and  d  the  driven 

pulley.  The  diameters  of  the  pulleys  can  be  of 
any  desired  proportions  to  suit  the  work  of  the 
machine.  ' 

D  =  diameter  and  R  =  radius  in  inches  of  the 

driving  pulley. 
d=  diameter  and  r  =  radius    of  the    driven 

pulley. 
L  =  length   and   5=  breadth   of   the   belt   in 

inches. 
F=  force  of  tension  in  pounds  of  the  pulling 

side  of  the  belt. 

f=  force  of  tension  on  the  slack  side. 
V=  velocity  of  the  belt  in  feet  per  second. 
8=  distance  in  inches  between  the  centres  of 
the  two  pulleys. 

JVand  n  =  numbers  of  revolutions  per  minute  of  the  respective  pul- 
leys D  and  d. 

<p  =  angle  in  degrees  occupied  by  the  belt  on  the  small  pulley. 
IP  =  horse-power  transmitted  by  the  belt. 

Revolutions  N :  n  =  d  :  D,  diameters.     The  revolutions  are  inverse 
as  the  diameters. 


ND 


ND 

* 
n 


The  force  bearing  in  the  journals  of  each  shaft  is  J?+f,  or  the  sum 
of  the  tensions  of  each  side  of  the  belt. 


BELT  AND  PULLEYS. 


97 


Fig.  82. 


The  force  which  transmits  the  motion  is  F  —  f,  or  the  difference 
between  the  two  tensions. 

The  effective  power  transmitted  is  equal  to  the  product,  of  the 
transmitting  force  and  the  velocity,  and  this  power  divided  by  550 
gives  the  horse-power. 

The  shafts  connected  by  belt  and  pulleys  need 
not  be  parallel  with  each  other,  but  they  must 
lay  in  parallel  planes,  as  represented  by  the  illus- 
tration, Fig.  82.  The  pulleys  must  be  placed  on 
the  shafts  so  that  the  driving  side  of  the  belt  forms 
right  angles  with  them.  The  belt  can  be  put  on 
so  as  to  drive  the  driven  pulley  in  any  desired 
direction. 

The  length  L  of  the  belt  will  be  found  .by  the 
following  formula : 


Fig.  83. 


L  =  TT  (E  +  r)  +  2  j/tf2  +  (  R  -  r}\ 

When  the  diameters  of  the  pulleys  are  alike,  or 
=  d,  the  length  of  the  belt  will  be 


The  shafts  or  the  belt  can  be  twisted  to  any  de- 
sired angle,  like  in  this  illustration,  Fig.  83,  the 
belt  is  twisted  180°,  and  the  driven  pulley  will 
then  run  in  an  opposite  direction  to  that  in  Fig.  81. 
The  belt  should  be  laid  on  so  as  to  have  the  same 
side  on  both  pulleys,  and  the  insides  will  then  rub 
flat  against  one  another  in  the  crossing  c. 

Slip  of  Belt. 


The  slip  of  belt  on  equal  pulleys  has  been  found 
by  experience  to  vary  between  2  and  3  per  cent,  under  ordinary  cir- 
cumstances. 

n  =  theoretical  revolutions  per  minute  of  the  driven  pulley. 
ri  =  actual  revolution  after  the  slip  is  deducted. 


98 


ELEMENTS  OF  MECHANICS. 


\  36.     FORMULAS   FOR   LEATHER  BELTS   ON   CAST-IRON   PULLEYS. 

For  Notation  of  Letters,  see  page  96. 


Force  and  Power  of  Trans- 


-re- 


126500 H> 


126500  IP 


DN(F-f) 
126500 

dn_DN 

230     230  ' 

126500  IP 
~  d(F-f)' 


Breadth  of  Belt  from  Expe- 
riments. 

,    2^/IP 


n  d  <f> 

2AF 

d   ' 

432  F 
d  <f> 

4320  IP 
=  ~n  d 

7.8  FV 
nd 

Bnd. 
"  4320 

_B_nd  a 

"    24 

Bd 
"  2.4'     ' 


10 
11 
12 

13 
14 

15 
16 


§37.     VULCANIZED    RUBBER   BELTS. 

The  vulcanized  rubber  belting  made  by  the  New  York  Belting  and 
Packing  Company  is  composed  of  heavy  cotton  duck,  woven  expressly 
for  that  purpose  and  vulcanized  between  layers  of  a*  metallic  alloy,  by 
which  process  the  stretch  is  entirely  taken  out  and  the  surface  made 
perfectly  smooth. 

This  belting  is  said  to  be  superior  to,  and  is  furnished  for  about 
half  the  price  of,  that  of  leather.  It  will  stand  a  heat  of  300°  Fahr., 
and  the  severest  cold  will  not  affect  its  good  quality,  even  if  run  in 
wet  places  or  exposed  to  damp  weather. 

The  friction  of  the  vulcanized  rubber  belting  is  about  double  that 
of  leather,  which  makes  it  less  liable  to  slip  on  the  pulley. 


ATTRACTION.  99 


§38.    MATTER. 

Matter  is  that  of  which  bodies  are  composed,  and  occupies  space. 

Matter  is  recognized  as  substance  in  contradistinction  from  geo«- 
metrical  quantities  and  physical  phenomena,  such  as  color,  shadow, 
light,  heat,  electricity  and  magnetism. 

We  have  no  knowledge  of  the  origin  or  source  of  matter,  but  only 
know  its  existence  and  obedience  to  forces.  Chemistry  has,  thus  far, 
dissolved  matter  into  some  sixty-five  distinct  elements,  but  in  the 
philosophy  of  mechanics  we  treat  matter  only  as  one  simple  element 
in  relation  to  the  three  physical  elements — -force,  'motion  and  time. 

These  four  elements — force  F,  motion  V,  time  T  and  mass  M— 
are  what  constitute  nature,  and  their  different  combinations  cause  the 
phenomena  which  we  study  and  observe. 

The  three  first  elements — F,  V  and  T—  are  what  constitute  life, 
which  physical  combination  with  matter  constitutes  organic  bodies. 

Physics  divides  matter  into  atoms,  molecules,  particles  and  bodies. 

Atom  is  the  ultimate  portion  into  which  matter  can -be  divided. 

Molecule  is  a  group  of  atoms. 

Particle  is  a  group  of  molecules. 

Body  is  a  group  of  particles,  consisting  of  molecules  and  atoms  of 
matter. 

§39     ATTRACTION  OR  GRAVITATION. 

It  is  a  well-known  and  established  fact  that  all  bodies  in  nature 
have  a  mutual  tendency  to  attract  each  other,  the  action  of  which  is 
called  universal  attraction  or  gravitation.  It  is  a  constant  action  be- 
tween all  kinds  of  matter,  which  cannot  be  disturbed  by  any  other 
cause. 

Attraction,  gravitation  and  gravity  mean  the  san^e  physical  action, 
but  there  exist  different  kinds  of  tendencies  between  different  kinds 
of  matter  to  attract  each  other,  which  are  independent  of  the  general 
law  of  gravitation ;  such  as  cohesion,  capillar,  molecular,  chemical 
affinity,  electric  and  magnetic  attractions  and  repulsions,  which  are 
not  called  gravitation,  and  which  actions  are  much  under  the  control 
of  human  skill.  The  force  of  cohesion  can  be  destroyed  by  a  superior 
force  crushing  or  tearing  the  body  to  pieces,  or  by  the  application  of 
heat. 

The  atoms  of  matter  are  acted  upon  by  two  opposite  forces — namely, 
molecular  attraction  and  repulsion — which  cause  bodies  to  exist  in 
three  aggregate  forms — namely,  solid,  liquid  and  gaseous,  according  to 
the  relation  between  the  two  forces. 


100  ELEMENTS  OF  MECHANICS. 

When  the  force  of  attractiou  is  superior  to  that  of  repulsion,  the 
body  will  have  the  consistency  of  a  solid ;  but  when  the  two  forces 
are  in  equilibrium,  the  body  will  have  a  liquid  form ;  and  when  the 
force  of  repulsion  is  superior  to  that  of  attraction,  the  body  will  con- 
sist in  the  form  of  a  gas. 

It  has  been  anticipated  that  matter  may  be  converted  into  a  fourth 
aggregate  form — namely,  that  of  an  imponderable  substance — but  the 
suggestion  is  not  realized  as  a  fact. 

All  bodies  in  nature  may  be,  or  are  capable  of  being,  converted  alter- 
nately into  the  three  aggregate  forms — namely,  solid,  liquid  and  gase- 
ous— although  we  have  not  yet  so  succeeded  with  some  bodies.  Ice, 
water  and  steam  are  the  three  aggregate  forms  of  one  body. 

It  appears  that  the  temperature  of  heat  is  the  force  of  repulsion, 
and  that  the  absence  of  heat  (cold)  allows  the  force  of  attraction  to 
draw  the  atoms  of  matter  in  closer  contact.  In  the  case  of  liquids, 
it  is  said  that  the. two  forces  are  in  equilibrium,  but  we  know  that 
liquids  differ  widely  in  temperature,  like  that  of  water  and  molten 
iron. 

If  the  force  of  repulsion  increases  in  some  ratio  with  the  tempera- 
ture, the  force  of  attraction  of  the  atoms  of  different  elements  must 
differ  in  accordance  with  the  force  of  repulsion,  for,  otherwise,  liquids 
could  not  vary  so  much  in  temperature. 

The  physical  constitution  of  the  force  of  attraction  is  a  mystery  to 
us,  and  we  have  yet  no  hope  of  its  ever  being  revealed.  If  there 
were  eye-bolts  in  each  atom  of  matter,  and  all  were  thus  connected 
by  elastic  strings  whose  elasticity  diminished  as  the  square  of  their 
length,  some  conception  could  be  formed  of  the  nature  of  attraction ; 
but  as  there  are  no  such  eye-bolts,  how  does  the  force  of  attraction 
take  hold  of  the  atom  ? 

It  is  self-evident,  however,  that  something  must  exist  between 
matter  to  form  the  connection  of  attraction  ;  and  whether  this  "  some- 
thing" be  elastic  strings  or  not,  it  cannot  be  cut  off  or  interfered  with 
in  the  least  by  any  intervening  means. 

Universal  attraction,  means  the  general  force  of  attraction  be- 
tween the  heavenly  bodies. 

Gravitation,  means  the  same  universal  force  of  attraction,  but 
implies  that  action  on  or  near  the  surface  of  the  earth  or  of  any  other 
heavenly  body. 


ATTRACTION.  101 


|40.     LAW    OF   ATTRACTION. 

The  law  of  universal  attraction  was  anticipated  by  Copernicus, 
Tycho  Brahe,  Kepler,  Fermat,  Roberval  and  Hook,  and  finally  estab- 
lished by  Sir  Isaac  Newton.  It  is  expressed  as  follows : 

The  force  of  attraction  is  directly  as  the  mass,  and  inversely  as  the 
sqttare  of  the  distance. 

This  expression  will  hold  good  when  the  mass  of  either  one  of  the 
attracting  bodies  is  taken  as  a  unit,  but,  more  correctly,  the  law  ought 
to  be  expressed  thus : 

The  force  of  attraction  between  any  two  bodies  is  equal  to  the  mass 
of  the  one  body,  'multiplied  by  that  of  the  ot/ier,  and  the  product  divided 
by  the  square  of  their  distance  apart. 

This  is  the  universal  law  of  attraction  or  gravitation  upon  which 
our  existence  wholly  depends. 


{41..    ILLUSTRATION    OF   THE   LAW   OF   ATTRACTION. 

Let  a  and  b,  Fig.  84,  represent  two  particles  of  matter,  supposed  to 
constitute  one  body,  and  c,  d,  e  and/  four  particles,  constituting  an- 
other body.  Each  of  the  six  particles  is  supposed  to  contain  one 
unit  of  matter,  and  the  distance  D  between  the  two  bodies  to  be  one 
unit  of  length. 

Draw  straight  lines  between  the  particles,  as  shown  in  the  illus- 
tration. 

The  particle  a  will  attract  the  particle  e,  as  well  as  d,  e  and/,  each 
with  one  unit  of  force ;  the  attraction, 
therefore,  between  the  particle  a  and 
the  body  cdef  will  be  four  units    of 
force ;  and  the  particle  b  will  also  at-  ?  . 
tract  the  body  cdef  with  four  units  of 
force ;   so  that  the  attraction   between 
the  bodies  a  b  and  cdef  will  be  eight 

units  of  force,  as   represented  by  the   eight  lines   drawn   between 
them. 

The  mass  of  the  body  a  b  is  2,  that  of  c  d  efis  4,  and  the  product 
of  2  and  4  is  8,  the  force  of  attraction,  according  to  Newton's  law. 
It  is  assumed  in  the  illustration  that  the  distance  between  the  bodies 
is  one  unit  of  length,  but  if  the  distance  be  two  units  of  length,  the 
force  of  attraction  will  be  only  2,  and  if  the  distance  D  is  only 
half  a  unit  of  length,  the  force  of  attraction  will  be  32. 


102  ELEMENTS  OF  MECHANICS. 


Let  F  denote  the  force  of  attraction  between  any  two  bodies  of 
masses  M  and  TO,  D  =  distance  between  the  bodies,  then 
,    Mm 


All  force,  power  and  work  are  derived  from  this  law. 
The  attraction  of  the  sun  draws  heavenly  bodies  into  it,  and  the  heat 
generated  by  the  collision  is  returned  into  space.  It  is  the  heat  and 
light  from  the  sun  which  decompose  carbonic  acid  in  our  atmosphere, 
and  promote  the  growth  of  vegetation  on  the  surface  of  the  earth,  by 
which  we  are  supplied  with  food  and  fuel  for  motive-power.  The 
burning  of  the  fuel  reproduces  carbonic  acid,  which  rises  into  the  air, 
where  it  is  again  decomposed  by  the  heat  of  the  sun.  Thus,  the  work 
of  heat  is  absorbed  and  reproduced  alternately  for  ever  and  ever. 

These  ideas  of  the  sources  of  work  agree  with  those  of  George 
Stephenson,  Sir  William  Thomson,  Waterston  and  Rankin. 

§  42.  Suppose  three  bodies  A,  B  and  C,  to  be  fixed  in  a  straight  line, 
Fig.  85.  and  their  distances  apart  as  repre- 

.«, >* £. 1,     sented  by  the  letters  a,  b  and  c. 

(j£\ /^       Let  the  mass  or  the  real  quan- 

VJ^  "j     tity  of  matter  in  each  body  be 

represented    by    its    letter,    say 
A  =  16,  B  =  8,  and  C=  4,  their  distances  apart  being  a  =  I,  b  =  2,  and 

Then  the  forces  of  attraction  between  the  bodies  will  be  as  follows : 

Between  A  and  B,  force  of  attraction  =  — —  = =  128. 

a  1 

B  C     8x4 

Between  B  and  C,  force  of  attraction  = = •  =  8. 

b*  4 

A  C     16x4 

Between  A  and  C,  force  of  attraction  = =• =  7.1. 

c*  9 

The  attraction  between  the  two  bodies  A  and  C  is  not  interfered 
with  by  the  body  B ;  nor  will  other  bodies  stationed  or  moving  be- 
tween or  about  either  one  or  both  the  masses  A  and  0  influence  the 
force  of  attraction  between  A  and  C.  Therefore,  the  force  of  attrac- 
tion between  any  two  particles  of  matter,  whether  embodied  in  a  solid 
or  porous  mass  or  isolated  in  empty  space,  is  equal  to  the  product  of 
the  masses  of  the  particles  divided  by  the  square  of  their  distance 
apart. 

§  43.  Suppose  a  particle  of  matter,  A,  Fig.  86,  to  be  enclosed  with- 
in a  hollow  material  sphere,  abed,  the  particle  being  a  unit  of 


ATTRACTION. 


103 


matter  acted  upon  in  all  directions  by  the  attraction  of  the  spherical 
surface. 

Draw  the  straight  lines  a  d  and  b  c  through  the  particle  A  to  re- 
present the  sides  of  the  two  cones  with  a  common 
vertex  at  A.  Let  8  represent  the  height  of  the 
large  cone,  and  r  that  of  the  small  one.  (a  6)  =  di- 
ameter and  (a  Z>)2  the  area  of  the  base  of  the 
small  cone  ;  (c  d)  =  diameter  and  (c  c?)*  the  area 
of  the  base  of  the  large  cone. 

The  particle  A  is  attracted  in  opposite  direc- 
tions by  the  matter  in  the  bases  of  the  cones 
which  constitute  parts  of  the  material  spherical 
surface. 


The  attraction  in  the  direction  of  the  arrow  r  is 


(at? 


In  the  direction  of  the  arrow  $  the  attraction  is 


The  angles  of  the  sides  of  the  two  cones  are  alike,  or  the  angles 
aAb  =  cAd.  .  • .  (a  b)  :  r  =  (c  d)  :  8. 


r  S   '  r*  S* 

That  is  to  say,  the  opposing  forces  of  attraction  are  alike  or  in 
equilibrium  at  the  particle  A,  or  that  the  particle  is  equally  attracted 
from  all  sides  by  the  spherical  surface. 

Now  let  the  spherical  shell  be  filled  up  with  matter  from  the  outer 
to  the  inner  dotted  concentric  circle  A ;  the  equilibrium  of  attraction 
on  the  particle  A  will  not  be  disturbed  by  that  matter. 

§  44.  If  the  particle  A  is  located  outside  of  the  spherical  shell,  as 
represented  by  Fig.  87,  the  formulas  of  attractions 
will  be  the  same  as  those  for  Fig.  86,  but  both  the  at- 
tractions will  in  this  case   act  in  one  and  the  same          T- 
direction  on  the  particle,  and  the  force  of  attraction 

will  be 


Fig.  87. 


Let  e  denote  the  distance  of  the  particle  from  the 
centre  of  the  sphere,  and  R  =  radius  of  the  sphere,  then 


e  =  r  +  R,         and  r  =  e  —  R. 

z  bf     (c  dY       (a  bY     (c  dY 
J    .  \       )  _    V       /     .  V       ) 


(e-R)*    e+R 


104  ELEMENTS  OF  MECHANICS. 

That  is  to  say,  if  the  matter  of  the  bases  of  the  cones  (a  6)2  and 
(c  c?)2  were  located  in  the  centre  of  the  sphere,  the  force  of  attraction 
of  that  matter  on  the  particle  A  would  be  the  same  as  when  at  the 
surface  of  the  sphere.  Therefore,  if  the  hollow  sphere  be  filled  up 
solid  with  matter,  its  force  of  attraction  on  the  particle  A  would  be 
equal  to  the  mass  of  that  matter  divided  by  the  square  of  the  distance 
from  the  centre  of  the  sphere  to  the  particle. 

§  45.  Fig.  88  represents  a  section  of  a  solid  sphere  of  homogeneous 
matter.     A  particle  at  A  is  attracted  toward  the  centre  only  by  the 
Fj    gg  concentric  sphere  enclosed  by  the  dotted  circle. 

All  attractions  of  the  hollow  sphere  outside  of  the 
dotted  circle  are  in  equilibrium  on  the  particle  A, 
as  proved  by  §  43. 

Let  R  denote  the  radius  of  the  whole  sphere, 
and  ?-  =  that  of  the  inner  sphere.  The  mass  or 
quantity  of  matter  in  a  sphere  can  be  represented 
by  jR*  or  r*.  Then  the  force  of  attraction  on  the 

i* 
particle  A  will  be  —  =  ?-. 

That  is  to  say,  the  force  of  attraction  on  any  particle  of  the  matter 
in  a  solid  sphere  is  proportionate  to  the  distance  r  of  that  particle 
from  the  centre. 

Suppose  a  hole  to  be  made  from  the  surface  to  the  centre  of  the 
earth,  and  a  body  let  down  into  it  by  a  rope  attached  to  a  balance- 
scale  ;  then  the  weight  of  the  body  will  decrease  with  the  depth  until 
it  reaches  the  centre  of  the  earth,  where  it  will  indicate  no  weight  on 
the  balance-scale  placed  at  the  surface,  omitting  the  weight  of  the 
rope. 

r  =  radius  of  the  earth. 

d  =  any  depth  to  which  the  body  is  sunk  in  the  hole. 
W=  weight  of  the  body  at  the  surface  of  the  earth,  and 
w  =  weight  of  the  body  at  the  depth  d. 


The  radius  r  and  depth  d  may  be  expressed  in  any  unit  of  length, 
as  well  as  W  and  w  in  any  unit  of  weight. 

The  mean  radius  of  the  earth  is  about  r  =  3956  miles,  or 
r  -  20887680  feet. 

When  r  and  d  are  expressed  in  feet,  the  weight  w  will  be 

W  (20887680  -d) 
20887680 


WEIGHT.  105 


WEIGHT. 

§  46.  The  weight  of  a  body  is  the  force  of  attraction  between  the 
earth  and  that  body.  The  weight  of  a  body  is  greatest  at  the  surface 
of  the  earth,  and  decreases  above  or  below  that  surface.  Above  the 
surface  the  weight  decreases  as  the  square  of  its  distance  from  the 
centre  of  the  earth,  and  below  the  surface  the  weight  decreases  simply 
as  its  distance  from  the  centre. 

The  weight  of  a  body  A,  Fig.  87,  weighing  100  pounds  at  the  sur- 
face a,  b  of  the  earth,  would  weigh  only  25  pounds  at  a  height  equal 
to  the  radius  of  the  earth  above  a  b. 

A  body  A,  Fig.  88,  weighing  100  pounds  at  the  surface  of  the 
earth,  would  weigh  only  50  pounds  at  a  depth  of  half  the  radius  be- 
low the  surface.  Therefore,  the  weight  of  a  body  is  not  a  constant 
quantity,  whilst  the  quantity  of  matter  in  the  body  or  the  mass  is 
constant  wherever  the  body  is  weighed. 

A  wholly  isolated  body  has  no  weight,  but  is  an  inert  mass,  in- 
capable within  itself  of  changing  its  own  motion  or  rest.  Any  change 
in  motion  or  rest  of  a  body  is  derived  from  external  force.  The 
weight  of  a  body  is  measured  by  the  pressure  it  produces  on  its  sup- 
port. Two  bodies  in  equilibrium  on  a  balance-scale  at  the  surface  of 
the  earth  will  also  be  in  equilibrium  above  or  below  that  surface,  be- 
cause the  force  of  gravity  acts  equally  on  both  bodies  ;  but  the  force 
supporting  the  balance-scale  varies  in  accordance  with  the  law  of 
gravity.  A  spring-balance  will  indicate  the  true  weight  of  a  body 
hung  upon  it  wherever  it  is  weighed. 

The  force  of  attraction  between  the  earth  and  one  gallon  of  distilled 
water  at  the  level  of  the  sea,  in  latitude  of  London,  51°  31'  N.,  is  10 
pounds  avoirdupois. 

The  standard  English  gallon  contains  277.274  cubic  inches.  The 
temperature  of  the  water  and  of  the  air  in  which  it  is  weighed  should 
be  62°  Fahr.,  and  the  barometer  30  inches. 

This  is,  however,  not  the  true  force  of  attraction  between  one  gal- 
lon of  water  and  the  earth,  for  we  must  add  the  weight  of  277.274 
cubic  inches  of  air  which  are  displaced  by  the  water.  The  weight  of 
a  cubic  foot  of  dry  air  of  temperature  62°  Fahr.  is  530  grains,  which 
will  be  0.01215  pounds  for  the  capacity  of  one  gallon.  Therefore, 
when  a  gallon  of  water  weighs  10  pounds  in  air  the  force  of  attrac- 
tion between  that  water  and  the  earth  will  be  10.01215  pounds. 


106  ELEMENTS  OF  MECHANICS. 


MASS. 

§  47.  Mass  is  the  real  quantity  of  matter  in  a  body,  and  is  pro- 
portioned to  weight  when  compared  in  one  or  the  same  locality. 
Mass  is  a  constant  quantity,  whilst  weight  varies  with  the  force  of 
gravity  which  produces  it. 

The  force  of  gravity  accelerates  or  increases  the  velocity  of  a  fall- 
ing body  at  the  rate  of  32  feet  per  second  at  the  surface  of  the  earth. 
This  velocity  is  called  the  acceleratrix  of  gravity,  and  is  generally  de- 
noted by  the  letter  g. 

When  a  force  acts  constantly  on  a  body  free  to  move,  and  the  di- 
rection of  the  force  passes  through  the  centre  of  gravity  of  the  body 
in  the  direction  of  motion,  the  velocity  of  the  body  will  increase  con- 
stantly as  long  as  the  force  acts  constantly. 

Let  M  denote  the  mass  of  a  body. 

F=  the  constant  acting  force. 

T=time  of  action. 

F=  velocity  of  the  body  at  the  end  of  the  time  T. 

These  quantities  bear  the  following  relation  to  each  other : 
M :  F=  T:  F,     and     MV=FT. 

These  functions  are  termed  dynamic  momentums,  and  distinguished 
as  follows : 

Momentum  of  motion  MV=  F  T,  momentum  of  time. 

Neither  of  these  momentums  should  be  termed  force. 

When  F  is  expressed  in  pounds,  T  in  seconds  and  F  in  feet  per 
second,  then  the  unit  of  mass  will  be  32.17  pounds,  which  is  equal  in 
number  to  the  acceleratrix  g  for  a  falling  body  at  the  surface  of  the 
earth. 

MATT. 

§  48.  No  specific  name  has  yet  been  given  to  any  unit  of  mass,  the 
want  of  which  makes  this  subject  somewhat  obscure.  Although  we 
are  told  that  mass  is  equal  to  the  weight  divided  by  the  acceleratrix 
<7  =  32.17,  it  does  not  make  the  same  impression  as  if  we  had  a  specific 
name  for  the  unit  of  mass,  for  which  reason  it  is  proposed  to  assign  a 
name  to  it — namely,  matt..,  from  the  word  matter ;  that  is  to  say,  one 
matt.  =  32.17  pounds,  or  the  mass  expressed  in  matts.  multiplied  by 
32.17  would  give  the  weight  of  the  mass  in  pounds. 

There  are  69.63  matts.  in  a  ton  weight  of  2240  pounds  of  matter. 

If  W  denotes  the  weight  of  a  body  in  pounds,  then  its  mass  ex- 
pressed in  matts.  will  be 

M-  W-    W 
J/~7~3217' 


MATT.  107 


One  matt.  =  the  mass  of  891  cubic  inches  of  distilled  water  of 
temperature  40°  Fahr. 

The  adoption  of  this  term  matt,  will  distinguish  mass  from  force. 
Although  the  weight  of  a  mass  is  force  of  gravity,  all  forces  are  not 
weights  of  matter.  The  force  which  sets  in  motion  a  railroad  train 
is  independent  of  the  force  of  gravity,  but  may  be  as  well  expressed 
by  weight,  as  the  mass  of  the  train ;  but  we  cannot  solve  the  dynami- 
cal action  without  converting  the  weight  or  mass  of  the  train  into 
matte. 

This  is  the  fundamental  principle  of  dynamics  of  matter,  which 
should  be  distinctly  understood  and  remembered,  and  is  of  so  great 
importance  that  it  is  well  worthy  of  repeating — namely, 

M:F=T:V,    and    HV=FT. 

We  have  heretofore  been  taught,  in  text-books  and  in  colleges,  that 
momentum  MV  is  force,  which  is  a  great  error. 

Force  is  only  one  element  of  momentum.  Dynamic  momentum 
divided  by  time  is  force ;  that  is  to  say,  if  a  mass  M=  4  matts.  moves 
with  a  velocity  of  V=  6  feet  per  second,  its  momentum  is  24.  If  a 
force  is  applied  to  stop  the  mass,  and  can  do  so  in  T=  3  seconds,  then 
the  force  of  that  momentum  is  24  :  3  =  8  pounds.  Mass  is  inert,  or 
incapable  of  changing  its  own  motion  or  rest,  and  can,  therefore,  not 
be  considered  as  force  whether  in  motion  or  at  rest. 

Force  is  required  in  bringing  a  mass  from  rest  to  motion,  or  from 
motion  to  rest ;  but  in  either  case  that  force  must  be  applied  from 
external  causes  independent  of  the  mass. 

When  a  body  in  motion  is  suddenly  stopped,  like  that  of  a  falling 
body  striking  the  ground,  it  is  stopped  by  the  force  of  resistance  it 
meets  with,  and  not  by  any  force  within  itself.  The  inertia  of  a  body 
free  to  move  presents  a  resistance  equal  to  any  force  applied  on  it, 
whether  in  motion  or  at  rest. 

1 49.     TABLE   FOR   THE   CONVERSION    OF   WEIGHT   AND   MASS. 

The  following  table  is  for  converting  weight  into  mass,  or  mass  into 
weight,  which  will  be  very  useful  in  examples  of  dynamics  of  matter. 
Example.  Weight  957  pounds  -  29.748  matts.     See  Table. 
Example.  Convert  the  mass  of  M  =  3466  matts.  into  pounds. 

Matts.  =  pounds. 

8460  =  111310 
6=   193 
Matts.  3466  =  111503  pounds. 


108 


CONVERSION  OF  WEIGHT  INTO  MASS. 


O 

1 

2 

TO 
3 

«TS  OP 

4 

POUNI 

5 

s. 
6 

7 

8 

9 

Lbs. 

matts. 

matts. 

matts. 

matts. 

matts. 

matts. 

matts. 

matts. 

matts. 

matts. 

0 

.03108 

.06217 

.09325 

.12434 

.15542 

.18651  .21759J.24868 

.27976 

10 

.31085 

.34193 

.37302 

.40410 

.43519 

.46627 

.49736  .52844!.  55953 

.59061 

20 

.62170 

.65278 

.68387 

.71495 

.74604 

.77712 

.80821 

.83929  '.87038 

.90146 

30 

.93254 

.96363 

.99472 

1.0258 

1.0569 

1.0880 

1.1191 

1.15011.1812 

1.2123 

40 

1.2434 

1.2745 

1.3056 

1.3467 

1.3678 

1.3989 

1.4300 

1.4610 

1.4921 

1.5232 

50 

1.5542 

1.5853 

1.6164 

1.6575 

1.6786 

1.7097 

1.7408 

1.7718 

1.8029 

1.8340 

60 

1.8651 

1.8962 

1.9273 

1.9684 

1.9895 

2.0206 

2.0517 

2.0827 

2.1138 

2.1449 

70 

2.1759 

2.2070 

2.2381 

2.2792 

2.3003 

2.3314 

2.3625 

2.3935 

2.4246 

2.4557 

80 

2.4868 

2.5179 

2.5490 

2.5801 

2.6112 

2.6423 

2.6734 

2.7044 

2.7355 

2.7666 

90 

2.7976 

2.8287 

2.8598 

2.8909 

2.9220 

2.9531 

2.9842 

3.0152 

3.0463 

3.0774 

100 

3.1085 

3.1396 

3.1707 

3.2018 

3.2329 

3.2640 

3.2951 

3.3261 

3.3572 

3.3883 

110 

3.4193 

3.4504 

3.4815 

3.5126 

3.5437 

3.5748 

3.6059 

3.6369 

3.6680 

3.6991 

120 

3.7202 

3.7612 

3.7924 

3.8235 

3.8546 

3.8856 

3.9168 

3.9478 

3.9789 

4.0100 

130 

4.0310 

4.0720 

4.1032 

4.1343 

4.1654 

4.1964 

4.2276 

4.2586 

4.2897 

4.3208 

140 

4.3419 

4.3829 

4.4141 

4.4452 

4.4763 

4.5073 

4.5385 

4.5695 

4.6006 

4.6317 

150 

4.6527 

4.6937 

4.7249 

4.7560 

4.7871 

4.8181 

4.8493 

4.8803 

4.9114 

4.9425 

160 

4.9636 

5.0046 

5.0358 

5.0669 

5.0980 

5.1290 

5.1602 

5.1912 

5.2223 

5.2533 

170 

5.2744 

5.3154 

5.3466 

5.3777 

5.4088 

5.4398 

5.4710 

5.5020 

5.5331 

5.5641 

180 

5.5853 

5.6263 

5.6575 

5.6886 

5.7197 

5.7507 

5.7819 

5.8129 

5.8440 

5.8750 

190 

5.8961 

5.9371 

5.9683 

5.9994 

6.0305 

6.0615 

6.0927 

6.1237 

6.1548  6.1858 

200 

6.2170 

6.2480 

6.2792 

6.3103 

6.3414 

6.3724 

6.4036 

6.4346 

6.4657!  6.4967 

210 

6.5278 

6.5588 

6.5900 

6.6211 

6.6522 

6.6832 

6.7144 

6.7454 

6.7765  6.8075 

220 

6.8387 

6.8697 

6.9009 

6.9320 

6.9631 

6.9941 

7.02537.0563 

7.0874 

7.1184 

230 

7.1495 

7.1805 

7.2117 

7.2428 

7.2739 

7.3049 

7.3361 

7.3671 

7.3982 

7.4292 

240 

7.4604 

7.4914 

7.5226 

7.5537 

7.5848 

7.6158 

7.6470 

7.6780 

7.7091 

7.7401 

250 

7.8612 

7.8022 

7.8334 

7.8645 

7.8956 

7.9266 

7.9578 

7.9888 

8.0199 

8.0509 

260 

8.1721J8.1131 

8.1443 

8.1754 

8.2065 

8.2375 

8.2687 

8.2997 

8.3308 

8.3618 

270 

8.3930'  8.4239 

8.4552 

8.4863 

8.5174 

8.5484 

8.5796 

8.6106 

8.6417 

8.6727 

280 

8.7038  8.7347 

8.7660 

8.7971 

8.8282 

8.8592 

8.8904 

8.9214 

8.9525 

8.9835 

300 

9.0146 
9.3255 

9.0455 
9.3564 

9.0768 
9.3877 

9.1079 
9.4188 

9.1390 
9.4499 

9.1700 
9.4809 

9.2012 
9.5121 

9.2322 
9.5431 

9.2633 
9.5742 

9.2943 
9.6052 

310 
320 

9.6363 
9.9472 

9.6672 
9.9781 

9.6985 
10.009 

9.7296 
10.040 

9.7607 
10.072 

9.7917 
10.102 

9.8229 
10.133 

9.8539 
10.165 

9.8850 
10.196 

9.9160 
10.227 

330  110.258 

10.289 

10.320 

10.351 

10.383 

10.413 

10.444 

10.476 

10.507 

10.538 

340 

10.569 

10.600 

10.631 

10.662 

10.694 

10.724 

10.755 

10.787 

10.818 

10.849 

350 

10.880 

10.911 

10.942 

10.973 

11.005 

11.035 

11.066 

11.098 

11.129 

11.160 

360 

11.191 

11.222 

11.253 

11.284 

11.316 

11.346 

11.377 

11.409 

11.440 

11.471 

3  7  Oil  1  1.501 

11.532 

11.563 

11.594 

11.626 

11.656 

11.687 

11.719 

11.750 

11.781 

380111.812 

11.843 

11.874 

11.905 

11.937 

11.967 

11.998 

12.030 

12.061 

12.092 

390 

12.123 

12.154 

12.185 

12.216 

12.248 

12.278 

12.309 

12.341 

12.372 

12.403 

400 

12.434 

12.465 

12.496 

12.527 

12.559 

12.589  12.620 

12.652 

12.683 

12.714 

410 

12.745 

12.776 

12.807 

12.838 

12.870 

12.900  12.931 

12.963 

12.994 

13.025 

420 

13.056 

13.087 

13.118 

13.149 

13.181 

13.211  113.242 

13.274!  13.305 

13.336 

430 

13.367 

13.398 

13.429 

13.460 

13.492 

13.522  113.553 

13.585 

13.616 

13.647 

440 

13.678 

13.709 

13.740 

13.771 

13.803 

13.833  13.864 

13.896 

13.927 

13.958 

450 

13.989- 

14.020 

14.051 

14.082 

14.113 

14.144 

14.175 

14.207 

14.238 

14.269 

460 

14.300 

14.331 

14.362 

14.393 

14.424 

14.455 

14.486 

14.518 

14.549 

14.580 

470 

14.610 

14.642 

14.673 

14.704 

14.735 

14.766  14.797  14.829  14.860  14.891 

480 
490 
500 

14.922  1  14.953  14.984  15.015  1  15.046 
15.232  15.264  15.295  15.396,  15.357 
15.5421  15.574  15.605!  15.6361  15.667 

15.077!  15  108 
15.388  15.419 
15.698  15.729 

15.140 
15.451 
15.760 

15.171 

15.482 
15.791 

15.202 
15.513 
15.823 

CONVERSION  OF  WEIGHT  INTO  MASS. 


109 


O 

1 

2 

TJ 

3 

1ITS  OF 

4 

potmr 
5 

>s. 
6 

7 

8 

9 

Lbs. 

matts. 

matts. 

matts. 

matts. 

matts. 

matts. 

matts. 

matts. 

matts. 

matts. 

500 

15.542 

15.574  15.605 

15.636 

15.667 

15.698 

15.729  15.760 

15.791 

15.823 

510 

15.852 

15.8841  15.915 

15.946 

15.977 

16.008'  16.039  16.070 

16.101 

16.133 

520 

15.162 

16.194  16.225 

16.256 

16.287 

16.318!  16.349  16.380 

16.411 

16.443 

530 
540 

16.473 
16.784 

16.505  16.536 
16.816  16.847 

16.566 
16.877 

16.598 
16.909 

16.6291  16.660 
16.940  16.971 

16.691  16.722  16.754 
17.002  '17.033  17.065 

550 

17.095 

17.127 

17.158 

17.188 

17.220 

17.251 

17.282 

17.313  17.344  17.376 

560 

17.406 

17.438 

17.469 

17.499 

17.531 

17.562 

17.593 

17.624 

17.655 

17.687 

570 
580 

17.717 

18.028 

17.748 
18.059 

17.779 
18.090 

17.810  17.842 
18.121  18.153 

17.872 
18.183 

17.903 
18.214 

17.934 
18.245 

17.966 
18.277 

17.997 
18.308 

590 

18.339 

18.370 

18.401 

18.4321  18.464 

18.494 

18.525 

18.556 

18.588 

18.619 

600 

18.651 

18.682 

18.712 

18.744 

18.775 

18.806 

18.837 

18.868 

18.899 

18.930 

610 

18.962 

18.993 

19.023 

19.055 

19.086 

19.117 

19.148 

19.179 

19.210 

19.241 

620 

19.273 

19.304 

19.334 

19.366  119.397 

19.428 

19.459 

19.490 

19.521 

19.552 

630 

19.584 

19.615 

19.646 

19.677  19.708 

19.739 

19.770 

19.801 

19.832 

19.863 

640 

19.895 

19.926 

19.957 

19.988 

20.019 

20.040 

20.081 

20.112 

20.143 

20.174 

650 

20.206  20.237  2.0.268 

20.299 

20.330 

20.351 

20.392 

20.423 

20.454 

20.485 

660 

20.517  20.548  20.579 

20.610 

20.641 

20.662 

20.703 

20.734 

20.765 

20.796 

670 

20.828  20.859 

20.890 

20.921 

20.952 

20.973 

21.014  21.045 

21.076 

21.107 

680 

21.139 

21.170 

21.201 

21.232 

21.263 

21.384 

21.32521.356 

21.387 

21.418 

690 

21.449 

21.480 

21.511 

21.542 

21.573 

21.694 

21.635121.666 

21.697 

21.728 

700 

21.  759  121.790 

21.821 

21.852 

21.883 

21.914 

21.945  21.976 

22.007 

22.038 

710  22.070  22.101 

22.132 

22.163 

22.194 

22.225 

22.256 

22.287 

22.318  22.349 

720  22.381!  22.41  2  22.443 
730,  22.692  22.723  22.754 

22.474  22.505 
32.785  22.816 

22.536 
22.847 

22.567  22.598 
22.878  22.909 

22.629 
22.940 

22.660 
22.971 

740 

23.003;  23.034 

23.065 

23.096  23.127 

23.158 

23.18923.220 

23.251  23.282 

750 

23.31423.34523.376 

23.407 

23.438 

23.469 

23.50023.531 

23.562  23.593 

760 

23.625  23.656 

23.687 

23.718 

23.749 

23.780 

23.811  23.842  23.873 

23.904 

770 

23.93623.967 

23.998 

24.029 

24.060 

24.091 

24.122  '24.153 

24.184!  24.215 

780 

24.246  24.277 

24.318 

24.339 

24.370 

24.401 

24.432  24.463  24.494  i  24.525 

790 

24.557  24.588 

24.629 

24.650 

24.681 

24.712 

24.743  24.774  24.805  24.836 

800 

24.868  1  24.899  24.930 

24.961 

24.992 

25.023 

25.054 

25.085  25.115  25.147 

810 

25.1791  25.210  25.241 

25.272 

25.303 

25.334 

25.365 

25.396 

25.427  25.458 

820 

25.49025.521 

25.5')2 

25.583 

25.614 

25.645 

25.676 

25.707  25.738:25.769 

830i!25.801  25.832  25.863 
840  126.112  26.143  26.174 
850  126.423  26.454  26.485 

25.894 
26.205 
26.516 

25.925 
26.236 
26.547 

25.956 
26.267 
26.578 

25.987 
26.298 
26.609 

26.018  26.049  126.080 
26.32926.36026.391 
26.640  26.671  !  26.7  02 

860, 

26.73426.765 

26.796 

26.827 

26.858 

26.889 

26.920 

26.951  26.982  127.013 

870 

27.045  27.076  '27.107 

27.138  27.169 

27.200 

27.231 

27.26227.29327.324 

880: 

27.356  27.387  27.418 

27.449 

27.480 

27.511 

27.542 

27.573  27.604  27.635 

890 

27.666  27.697 

27.728 

27.759 

27.790 

27.821  27.852  27.883 

28.914  27.945 

900 

27.976.28.008  28.039 

28.070 

28.101 

28.132 

28.163  28.194  28.225 

28.256 

910 

28.287  28.318  28.349 

28.380 

28.411 

28.442 

28.473  28.504  28.535  28.566 

920 

28.59828.62928.660 

28.691 

28.722 

28.753  28.784  28.815  28.846  28.877 

930 

28.90928.93028.971 

29.002 

29.033 

29.06429.095 

29.126  29.157  29.188 

940 

29.220  29.241 

29.282 

29.313 

29.344 

29.375-29.406 

29.437  !  29.468  '29.499 

950 

29.53129.55229.593 

29.624 

29.655 

29.686.29.717 

29.748  j  29.779  29.810 

960 

29.842^29.863  29.904  29.935 

29.966 

29.997  !  30.028  30.059  '  30.090  j  30.121 

970 

30.153  '  30.174 

30.215  30.246  30.277 

30  308  '30.339!  30.370  30.401  30.432 

980 

30.454  30.485  30.526 

30.557  30.588 

30.619  30.640  30.681:30.711  30.743 

990  30.765  30.796  30.837  30.868  j  30.899 
1000  31.085  31.107  31.148  31.179J31.210 

30.930'30.961!30.992  31.023  31.054 
31.241:31.272  31.303  31.334l31.365 

110 


CONVERSION  OF  MASS  INTO    WEIGHT. 


o 

1 

2 

UNITS  OF 

3         4 

MATTS. 

56         7         8 

9 

Malts     ibs. 

Ibs. 

Ibs. 

Ibs.        Ibs. 

Ibs.    |    Ibs.       Ibs.        Ibs. 

Ibs. 

0 

32.17 

64.34 

96.51  128.68 

160.85  193.02  225.19  257.36 

289.53 

lOi    321.7 

353.9 

386.0 

418.2    450.4 

4-32.5    514.7  !  546.9'  578.0 

611.2 

20    643.4 

675.6    707.7 

739.9    772.1 

804.2    836.4    868.6    899.7 

932.9 

30    965.1 

997.3  1029.4  1061.6  1093.8 

1125.9  1158.1  1190.3  1221.4 

1254.6 

40  1286.8 

1319.0;  1351.1  ;  1383.3  1415.5 

1447.6  1479.8  1512.0  1543.1 

1576.3 

50;  1608.5 

1640.7  1672.81  1705.0  1737.2 

1769.3  1801.5;  1833.7;  1864.8 

1898.0 

60!  ]  1930.2 

1962.4  1994.5  2026.7  2058.9 

2091.02123.22155.42186.52219.7 

70  2251.9 

2284.1  ;2316.2  2348.4  2380.6 

2412.7  2444.9  2477.1  25082  2541.4 

8012573.6 

2605  9  2637.9  2670.1  2702.3 

2734.4  2766.6  2798.9  2830.0;2863.1 

90IJ2895.3 

2927.6  2959.6  1  299  1.8  3024.0 

3056.1  3088.3  3120.6  3152.6  3184.8 

100  3217 

3249.3  3281.3  3313.5  3345.7 

3377.83410.0,3442.33474.3,3506.5 

110!3538.7 

3571.0  3603.0  3635.2  3667.  4 

3699.5  3731.7  3764.0  3796.0  3828.2 

120  3860.4 

3892.7;3924.7  3956.9  3989.1 

4021.2  4053.4  4085.7  4117.7 

4149.9 

130=4182.1 

4214.414246.4  4278.6  4310.8 

4342.9  4375.1  4407.4  4439.4I4471.6 

140  4503.8 

4536.0  4568.1  4600.3  4632.5 

4664.6  4696.8  4729.1  4761.1 

4793.3 

150;  4825.5 

48-37.7  4889.8  4922.0  4954.2 

4986.3  5018  5  5050.8  5082.8 

5115.0 

160  5147.2 

5179.4  5211.5  5243.7  J5275.9 

5308.0  5340.2  5372.5  5404.5 

5436.7 

170  5468.9 

5501.1  5533.2 

5565.4  5597.6 

5629.7  5661.9  5694.2  5726.2 

5758.4 

ISO  5790.6 
190'  61  12.3 

5822.8  5854.9  5887.1  5919.3 
6144.5  6176.6  6203.8  6241.0 

5951.4  5983.6  6015.9  6047.9  6080.1 
6273.1  6305.3  6337.61  6369.6  1  6401.  8 

200;  6434.0 

6466.2  6498.3  6530.5  6562.7 

6594.S  6627.0  6659.3  6691.3  6723.5 

210  6755.7 

6787.9  6820  0  6852.2  6884.4 

6916.5  6948.7  6981.0  7013.0  7045.2 

220|  7077.4 

7109.617141.7  J7173.9  7206.1 

7238.2  7270.4  7302.7  7334.7 

7366.9 

230  7399.1 

7531.3  !  7463.4  7495.6  7527.8 

7559.9  7492.1  7624.4  7656.4 

7688.6 

240  7720.8 

7753.0  7785.1 

7817.37849.5 

7881.6  7913.8  7946.1  7978.1 

8010.3 

250  8042.5 

8074.7  8106.8 

8139.08171.2 

8203.3  8235.5  8267.8  8299.8  8332.0 

260'  18364.2 

8396.4 

8428.5 

8460.7  8492.9 

8525.0  8557.2  !  8589.5  '8621.  5  8653.7 

270!  8685.9  8718.1 

8750.2 

8782.4  8814.6 

8846.7  8878.9  8911.1  8943.2 

8975.4 

2801  9007.6  9039.819071.9 

9104.119136.3 

9168.4  9200.6  9232.8  9264.9  9297.1 

290!  9329.3  9361.5  9393.6 
300  9651.0  9683.2  9715.3 

9425.8*9458.0 
9747.5  9779.7 

9490.1  !9522.3  9554.5  9586.6  9618.8 
9811.8,9844.0  9876.2  9908.3  9940.5 

310  9972.7  10005 

10037 

10069  10101 

10133.  10166  10198  10230 

10262 

320   10294 

10327 

10359 

10381   10423 

104551  10488   10520  10552 

10584 

330   10616  10649  10681 

10703!  10745 

10777   10810  10842  10874 

10906 

340   10938 

10970  11002 

11034!  11066 

11098!  11131   11163  11195 

11227 

350   11259  11292  11324 

11356  11388 

11420  11453  11485!  11517 

11549 

360   11581  11614  11646 

11678'  11710 

11742  11775  11807   11839   11871 

370   11903  11936 

11968 

12000|  12032 

12064  12097   12129  121611  12193 

380   12225  12257 

12289 

12321)  12353 

12385   12418   12450  12482 

12514 

390   12547   12-37<) 

15611 

12643   12675 

12707   12740   12772   12804 

12836 

400   12868 

12901 

12933 

12965!  12997 

13029   13062  13094  13126 

13158 

410  13189  13222 

13254 

13286  13318 

13350  13383:  13415  13447 

13479 

420   13510  13543  13575 

13607   13639 

13671   13704;  13736  13768 

13800 

430    13832  13865 

13897 

13929   13961 

13993  14026  14058  14090 

14122 

440   14154  14187 

14219 

14251    14283 

15315  14348   14380  11412 

14441 

450   14476   14508 

14540 

15572  14604 

14636  14669   14701   14733 

14765 

460   14798  14830 

14862 

148941  14926 

14958   14991   15023  15055 

15087 

470   15120  15152 

15184 

15216   15248 

15280  15313   15345  15377 

15409 

480  '15441  15473 

15505 

15537   15570 

15602   15634   15666   15698 

15731 

490   15763  15795   15827   15*59   15892 

15924  15956   15988  16020 

16053 

500   16085  16117   16149   16181   16214    16246   16278   16310  16342 

16375 

CONVERSION  OF  MASS  INTO    WEIGHT. 


Ill 


O 

1 

2 

XJ 
3 

NITS  01 

4 

-  MATT 

5 

3. 

6 

7 

8 

9 

Malts 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

500 

16085 

16117 

16149  16181 

16214 

16246  16278  16310 

16342 

16375 

510 

16406 

16438 

16470  16502 

16535 

16567  16599  16631  16663 

16696 

520  16728 

16760 

16792 

16824 

16857 

16889  16921  16953  16985 

17018 

530J 

17050 

17082 

17114 

17146 

17179 

17211 

17243  17275  17307 

17340 

540 

17372 

17404  17436 

17468 

17501 

175331  17565 

17597 

17629 

17662 

550 

17693 

17725 

17757 

17789 

17822 

17854 

17886 

17918 

17950 

17983 

560118015 

18047 

18079 

18111 

18144 

18176 

18208 

18240 

18272 

18305 

570i  18337 

18369 

18401 

18433 

18466 

18498 

18530 

18562 

18594 

18627 

580  18059 

18691 

18723 

18755 

18798 

18820 

18852 

18884 

18916 

18949 

590  18980 

19012 

19044 

19076 

19109 

19141 

19173 

19205 

19237 

19270 

600;  j  19302 

19334 

19366 

19398 

19431 

19463 

19495 

19527 

19559 

19592 

610  19624 

19656 

19688 

19720 

19753 

19785 

19817 

19849 

19881 

19914 

620  19946 

19978 

20010 

20042 

20075 

20107 

20139 

20171 

20203 

20236 

630  20-267 

20299 

20331 

20363 

20396 

20428 

20460 

20492 

20524 

20557 

640  |  20589 

20621 

20653 

20685 

20718 

20750 

20782 

20814 

20846 

20879 

650 

20911 

20943 

20975 

21007 

21040 

21072 

21104 

21136 

21168 

21201 

660 

21233 

21265 

21297 

21329 

21362 

21394 

21426 

21458 

21490 

21523 

670 

21554 

21586|  21618 

21650 

21683 

21715 

21747 

21779 

21811 

21844 

6SO 

21876 

21908 

21940 

21972 

22005 

22037 

22069 

22101 

22133 

22166 

690 

22198 

22230 

22262 

22294 

22327 

22359 

22391 

22423 

22455 

22488 

700  22519 

22551 

22583 

22615:  22648 

22680 

22712 

22744 

22776 

22809 

710  22841 

22873 

22905 

22937  22970 

23002 

23034 

23066 

23098 

23131 

7  20  '23103 

23194 

23227 

23259  23292 

23324 

23356 

23388 

23419 

23453 

730  234S4 

23516 

23548 

23580  23613 

23645 

23677 

23709 

23741 

23774 

740 

23806 

23838 

23S70 

23902  23935 

23967 

23999 

24031 

24064 

24096 

750!  24128 

241601  24192 

24224 

24257 

24289 

24321 

24353 

24386 

24418 

760'  24450 

24482 

24514 

24546 

24579 

24611 

24643 

24675 

24708 

24740 

770;  24771 

24803 

24835 

24867 

24900 

249321  24964 

24996 

25029 

25061 

780  i  25093 

25125 

25157 

25189 

25222 

25254  25286 

25318 

25350 

25383 

790  25415 

25447 

25479 

25511 

25544 

25576 

25608 

25640 

25671 

25705 

800 

25730 

25768 

25800 

25833 

25865 

25897 

25929 

25961 

25993 

26020 

S101  26057 

26089 

26121 

26154 

26186 

26218 

26250 

26282 

26314 

26347 

820  26379 

26411 

26443 

26476 

26508 

26540 

26572 

26604 

26636 

26669 

830  26701 

26733 

26765 

26798 

26830 

26862 

26894 

26926 

26958 

26991 

840!  27023 

'  27055 

27087 

27120 

27152 

27184 

27216 

27248 

27280 

27313 

850!  27344 

27376 

27408 

27441 

27473 

27505 

27537 

27569 

27601 

27634 

860 

27666 

27698 

27730 

27763 

27795 

27827 

27859 

27891 

27923 

27956 

870,  27988 

28020 

28052 

28085 

28117 

28149 

28181 

28213 

28245 

28278 

ssO  28309 

28341 

28373 

28406!  28438 

28470 

28502 

28534 

28566 

28599 

890  28031 

28603 

28695 

28728  28760 

28792 

28824 

28856 

28888 

28921 

900  28953 

.28985 

29017 

29050  29082 

29114 

29146 

29178 

29210 

29243 

910  29275 

29307 

29339 

29372;  29404 

29436 

29468 

29500 

29532 

29565 

920129590 

29628 

29660  29693  29725 

29757  29789 

29821 

29853 

29886 

930  29918 

29950 

29982  30015  30047 

30079  30111 

30143 

30175 

30208 

940  30240 

30272 

30304  !  30337  :  30369 

30401  30433 

30465 

30497 

30530 

950  30561 

30593 

30625'  30658  30690 

30722  30754  30786  30818 

30851 

960  30883 

30915 

30947  30980  31012 

31043  31076  31108  31140  31173 

970  31205 

31237 

31269  31302  31334 

31305  31398i  31430  31462  31495 

980  !  31527 

31559 

31591 

31624  31056 

31687  317201  31752  31784  31817 

990  :  31848 

31880 

31912 

31945  31977 

32009  32041  ;  32073  32105  32138 

1000  32170 

32202 

32234  32207  32299 

32331  32303  32395  32427  32400 

112  ELEMENTS  OF  MECHANICS. 


350.    THE   EARTH'S   ATTRACTION   ON    ITS   SURFACE. 

The  attraction  or  weight  of  bodies  on  the  earth's  surface  is  slightly 
influenced  by  various  causes,  namely  :  1st.  The  flatness  of  the  poles 
makes  the  radius  shortest  in  the  direction  of  the  earth's  axis,  which 
increases  the  force  of  attraction  at  the  poles. 

The  radius  H  of  the  earth  in  feet  at  any  latitude  L  is  about 

.#k  20887680(1  +  0.00164  cos.  2£). 

2d.  The  centrifugal  force  on  the  surface  of  the  earth,  which  varies 
as  the  cosine  for  the  latitude  and  acts  in  opposition  to  the  force  of 
attraction,  is  as  follows  :  ^ 

__  A  Rri1  cos.L*^" 

2933^5       ' 

in  which  A  =  unit  of  attraction  at  the  radius  R.     n  =  revolutions  per 
minute. 


(60x24)2    2073600 

3d.  The  centrifugal  force  of  the  earth's  rotation  around  the  sun 
influences  the  attraction  as  the  sine  of  the  sun's  angle  with  the  sur- 
face of  the  earth.  This  influence  is  so  insignificant  that  it  could  not 
be  detected  in  the  most  delicate  scientific  experiment  yet  known. 

4th.  The  attraction  of  the  sun  and  moon  influences  the  attraction 
of  bodies  on  the  earth's  surface  directly  as  the  sine  of  the  sun's  or 
moon's  angle  with  the  surface,  and  is  demonstrated  by  the  tidal  wave, 
causing  high  and  low  water. 

{51.     MASS   OF   THE    EARTH. 

It  is  not  necessary  for  the  purpose  of  this  treatise  to  observe  the 
precision  required  in  highly  scientific  investigations,  for  which  reason 
we  will  limit  ourselves  within  the  supposition  that  the  earth  is  a  per- 
fect sphere  of  radius  R  =  20,887,680  feet,  and  is  at  rest. 

The  delicate  pendulum  experiments  of  Cavendish  approximated  the 
mean  density  of  the  earth  to  be  5.44  times  that  of  water.  The  weight 
of  a  cubic  foot  of  distilled  water  of  temperature  39°  Fahr.  is  62.388 
pounds.  Then  the  mass  of  the  earth,  expressed  in  matts.,  will  be 

M^*&  62.388x5.44  =  402^35^00,000,000,000,000,000  matts. 
3  x  32.17 

Log.  M=  23.6050086. 


MASS  OF  TEE  EARTH.  113 

Let  r  denote  the  radius  of  the  earth  expressed  by  such  a  unit  of 
length  that  the  force  F  of  attraction  will  be 

82.17, 


which  is  the  force  of  attraction  between  the  earth  and  one  matt,  of 
matter  on  the  surface  of  the  earth. 

Then  we  have  that  radius  of  the  earth  to  be 

'          •'-li?'        and  r^lf, 111,886,700,000  units. 

Loff.r- 11.0487787. 
—  =  5356.59,        and  —  =  28,693,080. 


Log.  =  3.7288886.         Log.  =  7.4577772. 

The  force  of  attraction  F  in  pounds  between  any  two  masses  M 
and  m,  expressed  in  matts.,  and  their  distance  apart  D  in  feet,  will  be 


Example.  Two  bodies  J^=450  and  m  =  360  matts.  are  held  at  a 
distance  D  =  2  feet  apart.  Required  the  force  of  attraction  between 
them? 


^=2i52.  =  a0014115ofa^d- 

When  the  masses  of  the  attracting  bodies  are  expressed  by  weights 
W  and  w  in  pounds,  we  have 


W=  g  M,        and  w  =  g  m, 


W 
or  M  =  —  ,  and  m 


w 
—  . 

9  9 


y  y 

28693080  g1  =  29,694,700,000,        kg.  =  10.4726794. 
Force  of  attraction,  J-,  __|^__. 

10»  H 


114 


ELEMENTS  OF  MECHANICS. 


252.     ATTRACTION  OF   A   MOUNTAIN. 

The  accompanying  illustration  represents  a  mountain  of  known 
quantity  of  matter  M,  at  the  side  of  which  is  hung  on  a  string  a  b 
a  mass  m.  The  dotted  line  a  c  represents  the  vertical  in  which  the 
body  Ttt  would  hang  if  not  attracted  by  the  mountain,  and  a  b  is  the 
resultant  of  the  two  forces  of  attraction  from  the  earth  and  moun- 
tain. 

Let  v  denote  the  angle  b,  a,  c  formed  by  the  pendulum,  then  the 
cosine  for  v  represents  the  earth's  attraction,  and  sin.v  that  of  the 
mountain. 

W=  weight  of  the  body  m,  which  is  the  force  of  attraction  in  the 
direction  a,  c,  and  F—  force  of  attraction  of  the  mountain,  we  have 


Fig'89> 


F    sin.  v 


tan.v. 


W    cos.  v 
=  W  tan.v,     and      W^Fcot.v. 


To  find  the  distance  D  between  the  body  m  and  the  centre  of  at- 
traction of  the  mountain  is  a  rather  complicated  problem,  which, 
however,  can  be  approximated  by  knowing  the  form  of  the  moun- 
tain. 

With  the  body  m  as  a  centre  draw  concentric  circles  equal  dis- 
tances d  apart  through  the  mountain,  representing  spherical  sections 
of  areas  e,f,  g,  h,  etc.,  and  call  A  =  the  sum  of  all  the  spherical  sections. 

Reduce  each  area  to  a  zone  of  the  same  radius,  and  multiply  it  with 
cosine  for  }  the  angle  occupied  by  that  zone  ;  the  product  will  be  the 
areas  e,f,  g,  h,  to  be  inserted  in  the  following  formulas  : 


Then, 


and 


. 

f    g     h 
+•>  +  *  +  -  +,etc. 

\'l 


4     9    16 


The  base-line  of  the  mountain  need  not  be  referred  to  the  level  of 
the  sea,  but  to  the  level  of  the  ground  surrounding  it. 

Example  1.  The  quantity  of  matter  in  a  mountain  is  estimated  to 


INERTIA.  115 


be  M  =  1,250,000,000,000  matte.;  the  distance  between  the  mass  m  and 
centre  of  attraction  of  the  mountain  is  approximated  to  D  =  5000 
feet  ;  and  the  weight  of  the  body  TF=  321.7  or  the  mass  m  =  10 
inatts.  Required  the  force  of  attraction  between  the  body  m  and 
mountain  M  ? 

1,250,000,000,000x10 
==     28693080x5000- 

Example  2.  In  connection  with  the  preceding  example  the  body  m 
is  hung  on  a  string  20  feet  long.  Required  the  angle  v  and  the  linear 
deviation  a,  b  of  the  body  ? 


Formula,  tan.v  =  —  =          -  =  0.0000542, 
TV     321.7 

which  corresponds  to  an  angle  of  v  =  0°0'  11".     The  linear  deviation 
a,  b  will  be  a,  &  =  20  x«m.t;  =  0.0011  feet  =  0.0132  inches. 

An  artificial  horizon  at  the  place  of  the  body  m  would  have  the 
inclination. 


$  53.     INERTIA.  * 

Inertia  is  the  force  of  resistance  which  a  body  presents  to  any  ex- 
ternal force  tending  to  change  the  motion  or  rest  of  that  body. 

An  isolated  rigid  body  is  incapable  within  itself  of  changing  its 
own  state  of  motion  or  rest,  which  condition  is  called  inert. 

A  system  of  bodies  by  their  own  virtues  of  attraction  may  be 
capable  of  changing  each  other's  motion  or  rest,  as  is  the  case  with 
the  heavenly  bodies,  which  by  their  combined  attractions  regulate  one 
another's  motion  in  space. 

A  falling  body  does  not  move  simply  by  the  earth's  attraction 
alone,  but  by  the  combined  virtues  of  attraction  between  the  two 
bodies. 

If  a  1  ody  had  no  virtue  of  attraction,  no  other  body  could  exercise 
any  attraction  on  it,  but  there  exists  no  ponderable  matter  without  the 
virtue  of  universal  attraction. 

If  a  body  on  the  earth's  surface  could  be  deprived  of  its  virtue  of 
universal  attraction,  the  earth's  attraction  would  have  no  effect  upon 
it,  and  that  body  would  have  no  weight,  but  could  be  placed  at  rest 
anywhere  in  space  without  support. 

Two  bodies  of  iron  attract  one  another  by  their  virtue  of  magnetic 
attraction,  but  a  body  of  iron  exercises  no  magnetic  attraction  on  a, 
body  of  copper,  because  the  latter  possesses  no  such  virtue  of  attraction..' 


116  ELEMENTS  OF  MECHANICS. 


§54.     FALLING   BODIES. 

Falling  bodies  are  drawn  towards  the  earth  by  the  force  of  gravity 
pr  the  earth's  attraction,  which  force  is  equal  to  the  weight  of  the 
*ody. 

F:  M=  V:  T, 

But      F  =  W,    and    M=  —  , 


therefore  W  :  —  =  V  :  T  of  which  V=g  T. 

9 

That  is  to  say,  the  velocity  of  a  falling  body  in  feet  per  second  at 
the  end  of  fall  is  equal  to  the  time  of  fall  multiplied  by  the  accel- 
eratrix  g  =  32.17,  which  shows  that  the  body  gains  a  velocity  of  32.17 
feet  per  second  for  every  second  of  its  time  of  continued  fall. 

The  space  fallen  through  will  be  equal  to  the  time  of  fall,  multi- 

y 

plied  by  the  mean  velocity  —  . 
2 

Let  /S  denote  the  space  in  feet,  we  bave 

VT^g  T*  =  V*  =    V* 
2    ""    2        2g     64.34' 

By  solving  these  formulas  we  have  the  time  of  fall, 


=== 

g       V      \  g    =4.01' 
Let  u  denote  the  space  fallen  through  in  the  Tih  second,  we  have 

-       and    T=-+ 


£55.     FALL   OF   HEAVY   AND   LIGHT   BODIES. 

In  vacuum  a  heavy  body  does  not  fall  faster  than  a  light  one,  be- 
cause the  weight  of  each  body  is  equal  to  the  force  of  gravity  acting 
upon  it  ;  but  when  a  body  falls  in  air  or  in  a  liquid,  its  force  of  gravity 
is  diminished  equal  to  the  weight  of  the  air  or  liquid  displaced  by  the 
body  ;  and  whilst  the  mass  is  constant,  a  smaller  force  has  a  heavier 
body  to  move,  and  the  body  will  fall  slower.  A  pound  of  lead  dis- 


FALLING  BODIES. 


117 


places  less  weight  of  air  than  does  a  pound  of  cork,  for  which  reason 
the  lead  will  fall  faster  than  the  cork  in  air. 

The  force  of  gravity  must  also  overcome  the  resistance  of  the  air 
to  the  motion  of  the  falling  body,  which  is  independent  of  the  weight 
of  air  the  body  displaces.  This  resistance-  increases  as  the  square  of 
the  velocity  and  as  the  surface  exposed  to  the  motion.  A  pound  of 
cork  exposes  more  surface  to  the  motion  than  does  a  pound  of  lead, 
for  which  reason  the  cork  falls  much  slower. 

Formulas  with  these  influences  on  falling  bodies  will  be  given  far- 
ther on. 


£56.     ILLUSTRATION    FOR    FALLING    BODIES. 

The  illustration  represents  the  velocity  V,  space  /S  and  time  T  of  a 
body  falling  from  a  to  b,  with  tihe  corresponding  numbers  under  the 

respective  heads.    In  the 
second,   the 
from    T  to 


8 


Fig.:90. 


7  8  16 

9  10  25 

•9  12  36 

13  14  49 


15      16     64 


first 


body 
1",   a 

of  16.085  feet, 
which  is  represented  by 
tiie  first  triangle,  and 
which  is  the  unit  of  the 
-system.  The  velocity 
attained  in  this  period 
as  F=  2x16.085  =  32. 17 
feet  per  second.  If  the 
force  of  gravity  ceased 
'to  act  at  the  end  of  the 
first  second,  the  body 
would  continue  to  fall 
with  a  uniform  velocity 
of  32.17  feet  per  second ; 
but  as  the  force  -of  gravity  acts  constantly,  the  "body  will  attain  an 
additional  velocity  of  32.17  feet  for  every  second  of  its  fall. 

Between  1"  and  2"  the  body  passes  3  triangles,  which  is  the  space 
3  x  16.085  -  48.255  feet  passed  through  in  the  second  second  ;  and  the. 
velocity  at  2"  is  4  x  16.085  =  64.34  feet  per  second. 

In  the  eighth  second  of  its  fall  it  passes  15  triangles,  which  corre- 
spond to  15  x  16.085  =  241.275  feet  between  7"  and  8"  ;  and  the  ve- 
locity at  b  is  16  x  16.085  =  257.36  feet  per  second. 

From  a  to  b  the  body  passed  64  triangles,  which  represents  the 
linear  space  from  a  to  b,  or  64  x  16.085  =  1029.44  feet. 


118 


ELEMENTS  OF  MECHANICS. 


FALLING    BODIES. 

V=  velocity  in  feet  per  second  at  the  end  of  fall. 
T=  time  in  seconds  of  the  fall. 
S=  space  fallen  through  in  feet. 


V 

T 

S 

V 

T 

S 

V 

T 

8 

0.1 

0.0031 

.00015 

5.1 

0.1585 

0.4042 

11 

0.3419 

1.8804 

0.2 

0.0062 

.00031 

5.2 

0.1616 

0.4202 

12 

0.3730 

2.2380 

0.3 

0.0093 

0.0014 

5.3 

0.1647 

0.4364 

13 

0.4041 

2.6266 

0.4 

0.0124 

0.0025 

5.4 

0.1678 

0.4530 

14 

0.4352 

3.0464 

0.5 

0.0155 

0.0039 

5.5 

0.1709 

0.4700 

15 

0.4663 

3.4975 

0.6 

0.0186 

0.0055 

5.6 

0.1740 

0.4872 

16 

0.4973 

3.9784 

0.7 

0.0217 

0.0076 

5.7 

0.1771 

0.5047 

17 

0.5284 

4.4914 

0.8 

0.0248 

0.0099 

5.8 

0.1802 

0.5226 

18 

0.5595 

5.0355 

0.9 

0.0279 

0.0125 

5.9 

0.1833 

0.5407 

19 

0.5906 

5.6107 

1. 

0.0311 

0.0155 

6. 

0.1865 

0.5595 

20 

0.6217 

6.2170 

1.1 

0.0342 

0.0188 

6.1 

0.1896 

0.5782 

21 

0.6527 

6.8502 

1.2 

0.0373 

0.0224 

6.2 

0.1927 

0.5973 

22 

0.6838 

7.5218 

1.3 

0.0404 

0.0262 

6.3 

0.1958 

0.6168 

23 

0.7149 

8.2213 

1.4 

0.0435 

0.0304 

6.4 

0.1989 

0.6365 

24 

0.7460 

8.9520 

1.5 

0.0446 

0.0335 

6.5 

0.2020 

0.6565 

25 

0.7771 

9.7125 

1.6 

0,0477 

0.0381 

6.6 

0.2051 

0.6768 

26 

0.8082 

10.566 

1.7 

0.0508 

0.0432 

6.7 

0.2082 

0.6975 

27 

0.8393 

11.330 

1.8 

0.0539 

0.0485 

6.8 

0.2113 

0.7184 

28 

0.8704 

12.185 

1.9 

0.0580 

0.0551 

6.9 

0.2144 

0.7397 

29 

0.9015 

13.072 

2. 

0.0622 

0.0622 

7. 

0.2176 

0.7616 

30 

0.9325 

13.987 

2.1 

0.0653 

0.0685 

7.1 

0.2207 

0.7835 

31 

0.9636 

14.936 

22 

0.0684 

0.0756 

7.2 

0.2238 

0.8057 

32 

0.9947 

15.915 

±3 

0.0715 

0.0822 

7.3 

0.2269 

0.8282 

33 

1.0258 

16.926 

2.4 

0.0746 

0.0895 

7.4 

0.2300 

0.8510 

34 

1.0569 

17.967 

2.5 

0.0777 

0.0971 

7.5 

0.2331 

0.8741 

35 

1.0879 

19.038 

2.6 

0.0808 

0.1050 

7.6 

0.2362 

0.8975 

36 

1.1190 

20.142 

2.7 

0.0839 

0.1135 

7.7 

0.2393 

0.9213 

37 

1.1501 

21.277 

2.8 

0.0870 

0.1218 

7.8 

0.2424 

0.9453 

38 

1.1812 

22.443 

2.9 

0.0901 

0.1305 

7.9 

0.2455 

0.9697 

39 

1.2123 

23.640 

3. 

0.0932 

0.1398 

8. 

0.2487 

0.9948 

40 

1.2434 

24.868 

3.1 

0.0963 

0.1492 

8.1 

0.2518 

1.0168 

41 

1.2745 

26.127 

3.2 

0.0994 

0.1590 

8.2 

0.2549 

1.0451 

42 

1.3056 

27.417 

3.3 

0.1025 

0.1691 

8.3 

0.2580 

1.0707 

43 

1.3367 

28.739 

3.4 

0.1054 

0.1795 

8.4 

0.2611 

1.0966 

44 

1.3678 

29.407 

3.5 

0.1087 

0.1886 

8.5 

0.2642 

1.1228 

45 

1.3989 

31.475 

3.6 

0.1118 

0.2012 

8.6 

0.2673 

1.1494 

46 

1.4300 

32.890 

3.7 

0.1149 

0.2125 

8.7 

0.2704 

1.1762 

47 

1.4611 

34.330 

3.8 

0.1170 

0.2223 

8.8 

0.2735 

1.2034 

48 

1.4922 

35.813 

3.9 

0.1201 

0.2355 

8.9 

0.2766 

1.2259 

49 

1.5233 

37.321 

4. 

0.1243 

0.2486 

9. 

0.2797 

1.2586 

50 

1.5544 

38.830 

4.1 

0.1274 

0.2611 

9.1 

0.2828 

1.2867 

51 

1.5854 

40.413 

4.2 

0.1305 

0.2740 

9.2 

0.2859 

1.3151 

52 

1.6165 

42.029 

4.3 

0.1336 

0.2872 

9.3 

0.2890 

1.3438 

53 

1.6475 

43.659 

4.4 

0.1367 

0.2939 

9.4 

0.2921 

1.3729 

54 

1.6786 

45.322 

4.5 

0.1398 

0.3145 

9.5 

0.2952 

1.4022 

55 

1.7097 

47.017 

4.6 

0.1429 

0.3286 

9.6 

0.2983 

1.4318 

56 

1.7407 

48.740 

4.7 

0.1460 

0.3431 

9.7 

0.3014 

1.4618 

57 

1.7718 

50.396 

4.8   0.1491 

0.3578 

9.8 

0.3045 

1.4920 

58 

1.8029 

52.284 

4.9 

0.1522 

0.3729 

9.9  |  0.3076 

1.5226 

59 

1.8340 

54.103 

5. 

0.1554 

0.3885 

10.  1  0.3108 

1.5540 

60 

1.8651 

55.953 

FALLING  BODIES. 


119 


FALLING  BODIES. 

V=  velocity  in  feet  per  second  at  the  end  of  fall. 
T=  time  in  seconds  of  the  fall. 
8=  space  fallen  through  in  feet. 

V 

T 

S 

V 

T 

S 

V 

T 

8 

65 

2.0206 

65.669 

530 

16.478 

4366.6 

1030 

32.027 

16494 

70 

2.1769 

76.260 

540 

16.788 

4452.8 

1040 

32.338 

16815 

75 

2.3314 

87.427 

550 

17.099 

4701.7 

1050 

32.649 

17141 

80 

2.4868 

97.472 

560 

17.409 

4874.5 

1060 

32.950 

17463 

85 

2.6422 

112.29 

570 

17.720 

5050.2 

1070 

33.261 

17794 

90 

2.7976 

125.89 

580 

18.030 

5228.7 

1080 

33.572 

18129 

95 

2.9530 

140.27 

590 

18.341 

5410.6 

1090 

33.883 

18446 

100 

3.1085 

155.42 

600 

18.651 

5595.3 

1100 

34.194 

18806 

110 

3.4194 

188.07 

610 

18.961 

5783.1 

1110 

34.504 

19149 

1-20 

3.7302 

223.81 

620 

19.271 

5974.0 

1120 

34.815 

19496 

130 

4.0411 

262.67 

630 

19.582 

6168.3 

1130 

35.126 

19846 

140 

4.3519 

304.63 

640 

19.893 

6365.7 

1140 

35.436 

20198 

150 

4.6627 

349.70 

650 

20.204 

6566.3 

1150 

35.747 

20504 

160 

4.9736 

397.88 

660 

20.515 

6770.0 

1160 

36.058 

20913 

170 

5.2844 

449.18 

670 

20.826 

6976.7 

1170 

36.369 

21275 

180 

5.5953 

503.36 

680 

21.137 

7186.6 

1180 

36.680 

21641 

190 

5.9061 

561.08 

690 

21.448 

7399.5 

1190 

36.991 

22009 

200 

6.2170 

621.70 

700 

21.759 

7615.6 

1200 

37.302 

22381 

210 

6.5279 

689.43 

710 

22.070 

7834.8 

1210 

37.613 

22755 

220 

6.8387 

752.26 

720 

22.380 

8056.8 

1220 

37.924 

23133 

230 

7.1496 

822.20 

730 

22.691 

8282.2 

1230 

38.235 

23514 

240 

7.4604 

895.25 

740 

23.002 

8510.7 

1240 

38.546 

23898 

250 

7.7713 

971.41 

750 

23.313 

8742.4 

1250 

38.857 

24285 

260 

8.0821 

1050.6 

760 

23.623 

8976.7 

1260 

39.168 

24676 

270 

8.3930 

1133.1 

770 

23.934 

9214.6 

1270 

39.479 

25069 

280 

8.7038 

1218.5 

780 

24.245 

9455.5 

1280 

39.780 

25459 

290 

9.0147 

1308.2 

790 

24.556 

9699.6 

1290 

40.090 

25855 

300 

9.3255 

1398.8 

800 

24.868 

9947.2 

1300 

40.411 

26267 

310 

9.6363 

1493.7 

810 

25.179 

10197 

1310 

40.722 

26673 

320 

9.9472 

1591.6 

820 

25.490 

10451 

1320 

41.033 

27081 

330 

10.258 

1690.6 

830 

25.801 

10707 

1330 

41.343 

27493 

340 

10.569 

1791.7 

840 

26.112 

10967 

1340 

41.654 

27908 

350 

10.879 

1903.8 

850 

26.423 

11230 

1350 

41.965 

28326 

360 

11.190 

2014.2 

860 

26.733 

11495 

1360 

42.276 

28747 

370 

11.501 

2127.7 

870 

27.044 

11764 

1370 

42.587 

29172 

380 

11.812 

2244.3 

880 

27.354 

12035 

1380 

42.897 

29599 

390 

12.123 

2364.0 

890 

27.665 

12311 

1390 

43.208 

30029 

400 

12.434 

2486.8 

900 

27.976 

12589 

1400 

43.519 

30463 

410 

12.745 

2612.7 

910 

28.287 

12871 

1410 

43.820 

30893 

420 

13.055 

2741.5 

920 

28.598 

13155 

1420 

44.131 

31333 

430 

13.366 

2873.7 

930 

28.908 

13442 

1430 

44.442 

31776 

440 

13.677 

3008.9 

940 

29.219 

13733 

1440 

44.753 

32222 

450 

13.989 

3144.8 

950 

29.530 

14027 

1450 

45.064 

32671 

460 

14.300 

3289.0 

960 

29.841 

14323 

1460 

45.375 

33123 

470 

14.611 

3433.6 

970 

30.152 

14623 

1470 

45.686 

33579 

480 

14.922 

3581.3 

980 

30.463 

14927 

1480 

45.997 

34037 

490 

15.233 

3732.1 

990 

30.774 

15233 

1490 

46.308 

34499 

500 

15.545 

3886.2 

1000 

31.085 

15542 

1500 

46.631 

34973 

510 

15.856 

4043.3 

1010 

31.396 

15855 

1510 

46.732 

35082 

520 

16.167 

4203.4 

1020 

31.707 

16179 

1520 

47.043 

35752 

120 


ELEMENTS  OF  MECHANICS. 


£57.     FORMULAS   FOR   BODIES   FALLING    FREELY   UNDER   THE 
ACTION   OF   GRAVITY. 


Velocity   in    feet   per 
second  at  the  end  of 
fall. 

Space    in    feet  fallen 
through      in       the 
time  T. 

Time  of  fall  in  seconds. 

Space    in    feet  fallen 
through  in  the  Tth 
second. 

.     v. 

8. 

T. 

U. 

V=gT.         .1 

8=  9  ?\         5 

T=—.         .    9 

u  =  g(T-$).  13 

2 

9 

V-2-j->        -2 

S-^'       '  6 

T=  —  .       .  10 

T=-  +  $.      .  14 

F=^.3 

*"5'         '7 

Hy--11 

o         V* 

v^s 

F=8.021/S..4 

64.34 

4.01 

Example.  A  body  is  dropped  from  a  height  of  $=80  feet.     Ke- 
quired  its  time  of  fall,  and  with  what  velocity  it  reaches  the  ground? 


Time,  Formula  11.         T= 


2x; 


32.17 


=  5  seconds. 


Velocity,  Formula  4.      F=  8.02j/80  =  71.73  feet  per  second. 

£  58.     ASCENDING    BODIES. 

Ascending  bodies  are  raised  by  external  force  superior  to  that  of 
gravity.  The  start  of  a  body  from  rest  to  a  uniform  ascending  motion 
requires  a  greater  force  than  that  of  gravity  or  weight  of  the  body. 
The  actual  force  which  starts  the  body  vertically  upward  is  the  dif- 
ference between  the  applied  force  and  the  weight  of  the  body. 

The  force  required  to  maintain  a  uniform  ascending  motion  is 
equal  to  the  weight  of  the  body.  If  the  force  suddenly  ceases  to  act, 
the  body  will  continue  to  ascend  against  the  force  of  gravity  with  a 
retarded  motion  until  it  stops,  when  the  force  of  gravity  returns  the 
body,  which  is  then  said  to  fall. 

The  retarded  motion  of  a  body  ascending  against  the  force  of 
gravity  is  inversely  the  same  as  the  accelerated  motion  in  its  descent. 

Suppose  the  case  of  firing  a  ball  vertically  upward  from  a  gun : 


ASCENDING  BODIES. 


121 


F=  the  velocity  given  to  the  ball  at  the  muzzle. 

•y  =  the   velocity  at  the  time  t  from  the  time  the  ball  left  the 
muzzle  of  the  gun. 

T=  time  the  ball  will  ascend. 

t  =  any  time  less  than  T. 

S=  height  in  feet  to  which  the  ball  will  ascend. 

s  =  the  height  the  ball  ascends  in  the  time  t. 
In  the  accompanying  Fig.  91  a  represents  the  muzzle  of      Fig.  91. 
a  gun,  at  which  the  charge  of  powder  has  given  an  ascend- 
ing velocity  V  to  the  ball.     The  height  to  which  the  ball 
will  reach  at  c  is  found  by  Formula  7, 

s-?-. 


The  time  T  required  for  the  ascent  will  be  Formula  9, 


On  reaching  c  the  force  of  gravity  will  return  the  ball  to 
the  muzzle  a  in  an  equal  time  T  and  with  the  same  velocity 
V,  so  that  the  ascending  retarded  motion  is  inversely  the 
same  as  the  accelerated  descending  motion,  or  at  equal 
heights  above  the  muzzle,  the  ball  will  have  equal  velocities 
of  ascension  or  descension. 

If  t  denotes  the  time  in  which  the  body  ascends  from  a  to  b,  or 
descends  from  b  to  a,  and  T  the  time  from  a  to  <?,  or  from  c  to  a,  then 
the  time  between  a  and  c  will  be  T- 1,  and  the  space  or  vertical 
height  between  b  and  c  is  8-s.  Insert  these  values  in  Formula  11, 
which  will  be 


Formula  9, 


Then  we  have 


Z-,_J 

9  \ 


122  ELEMENTS  OF  MECHANICS. 

This  formula  gives  the  time  of  ascension  of  the  ball  from  a  to  b,  or 
descension  from  b  to  a  when  falling  from  c. 
The  velocity  v  of  the  ball  at  b  will  be 


Formula  3,  v  =-  j/2  g  (8-  s). 


=~    and    v^ 


from  which  v  =  l/V'i-2gs.     ...         15 

$59.     FORMULAS   FOR   BODIES   ASCENDING    UNDER    FREE  ACTION 
OF   GRAVITY. 


v  =  v/V1-2gs.          .         15 
Velocity  v  in  feet  per  second 

at  the  time  t  or  at  the  height  s.  v  =  -  -  ^-L  16 

t       2  ' 


Height  8  in  feet  at  the  time  t  2 

or  when  the  velocity  is  v.  a  ^ 

s  =  tv+y—.        .         .         18 


The  initial  velocity  Fin  feet  V=v+g  t.    .         .         .         19 
per  second  with  which  the  body 

starts  to  ascend  under  action  of  y=-  +  — .    ...         20 
gravity. 


t_-  21 

Time  t  in  seconds  required  to  g 

reach  the  height  s,  or  the  ve-  _  _  _  _ 

locity  reduced  to  v.  y  =  J^_    /J^1_J^£-  22 

~  g     v  91      9' 

The  formulas  for  T  and  8  are  the  same  as  those  for  falling  bodies. 

Example.  A  body  is  forced  vertically  upward  with  a  velocity  of 
V=  560  feet  per  second.  What  will  be  its  velocity  at  a  height  of 
s  =  420  feet  ?  and  to  what  height  will  it  ascend  ? 


Formula  15,  v  =  v/560*  -  2  x  32.17  x  420  =  535.25  feet  per  second, 
the  velocity  required. 

560* 

Formula  7,  S=  -  =  4874.1  feet,  the  height  required. 
2  x  32.17 


PARABOLIC  MOTION  OF  BODIES. 


123 


Fig.  92. 


|  60.     PARABOLIC  MOTION  OF  BODIES. 

A  body  B  thrown  up  from  a  in  the  inclined  direction  a,  I,  c,  d,  e 
will  describe  a  parabolic  curve  a,  b" ,  c",  d" ,  e,  as  represented  by  the 
illustration. 

z  =  b  a  b',  the  angle  of  inclination. 
V=  velocity  of  the  body  when  at  a, 
which  can  be   dissolved   into 
two  component  parts — namely, 
v  =  the  horizontal  velocity  with  which 
the  body  would  move  from  a 
to  b'  in  the  same  time  it  moves 
from  a  to  b,  and 

v'  =  vertical  velocity  with  which  the 
body  would  move  from  b'  to  b 
in  the  same  time  it  moves  from 
a  to  b. 

The  horizontal  velocity  v  will  be  uni- 
form, because  no  force  is  acting  9n  the 
body  in  that  direction ;  but  in  the  ver- 
tical direction  the  force  of  gravity  is 
constantly  acting  to  draw  the  body- 
down,  so  that  instead  of  arriving  at  b 
it  will  arrive  at  b",  or  the  force  of 
gravity  has  drawn  the  body  from  b  to  6";  thus,  the  vertical  velocity 
is  retarded  until  the  body  arrives  at  the  vertex  <?"  of  the  parabola, 
from  which  point  that  velocity  will  be  accelerated  until  the  body  ar- 
rives at  e  with  the  same  velocity  as  that  with  which  it  started  from  a. 
Without  the  action  of  the  force  of  gravity  the  body  would  continue 
with  a  uniform  velocity  V  via  b,  c,  d  to  e ;  but  the  force  of  gravity 
causes  the  body  to  fall 

1  unit  of  space  from  b  to  b" . 
4  units  of  space  from  c  to  c" '. 
9  units  of  space  from  d  to  d" . 
16  units  of  space  from  e  to  (f . 

The  space  fallen  through  is  as  the  square  of  the  time  of  fall.  The 
divisions  on  the  dotted  lines  represent  the  units  of  space  fallen 
through. 

"The  horizontal  constant  velocity,     v  =  Vcos.z. 
The  vertical  velocity  at  a,  v'  =  Vsin.z. 


124  ELEMENTS  OF  MECHANICS. 

T=  time  in  seconds  occupied  by  the  body  from  a  to  the  vertex  c'' 
of  the  parabola,  or  from  c"  to  d . 

—_  t^_  V  sin.z 

9          9 

t  =  any  time  in  seconds  from  a,  but  less  than  T. 
v"  =  vertical  variable  velocity  at  the  time  t. 

v"  =v'-gt  =  V sin.z- g  t. 

When  V  sin.z  =  g  T  the  vertical  velocity  will  be  0. 
V  =  velocity  in  the  direction  of  motion  at  any  time  t. 

The  height  h  of  the  vertex  e"  to  which  the  body  will  ascend  is 


R  =  base  of  the  parabola  which  is  generated  by  the  uniform  hor- 
izontal velocity  v  and  time  2  T.     R  is  called  the  horizontal  range. 

R  =  2v  T=2  TVCO&.Z. 
R         2  V*in.z 


2T= 
Horizontal  rajige,  R  = 


V  cos.z  g 

2  V1  sin.z  cos.z 


9 

With  a  definite  velocity  V  the  horizontal  range  R  is  greatest  when 
the  angle  z  =  45°,  because  the  product  of  sine  and  cosine  is  greatest 
at  that  angle. 

The  body  will  arrive  at  e  with  a  velocity  equal  to  that  with  which 
it  started  from  a,  omitting  the  resistance  of  the  air.  The  parabola 
a,  b",  c" ',  d",  e'  is  described  in  the  same  length  of  time  as  that  in 
which  it  would  fall  from  e  to  /, 

The  length  L  of  the  parabola  a,  b",  c",  d",  e"  is 

L~ 

Example.  A  body  is  thrown  up  with  a  velocity  V=  1000  feet  per 
second,  and  at  an  angle  z  =  60°.  Required  the  time  T,  velocities  v 
and  V ,  the  height  h  and  the  horizontal  range  Rt 

V  sin.z      1000x«m.60° 
Time  T= —— =  26.92  seconds. 


MOTION  ON  INCLINED  PLANES. 


125 


Fig.  93. 


The  body  will  complete  the  parabola  in  26.92x2  =  53.84  seconds. 
The  velocity  V  in  t  =  13.46  seconds,  will  be 

V  =  ^(lOOO  x  sin.&Q  -  32.17  x  13.46)'+ (1000  cos.60°)'  =  661.43 
feet  per  second. 

Horizontal  velocity  v  =  500  feet  per  second. 

The  height  i.Q^ffi.11667  fee, 

Horizontal  range JR  =  2vT=2x500x 26.92  =  26920  feet. 

£61.     MOTION   OF   BODIES  ON   INCLINED   PLANES. 

On  the  inclined  plane  I,  with  the  base  b  and  height  h,  is  placed 
body  or  mass  M.  Draw  from  the  centre 
of  M  the  vertical  line  W,  to  represent  the 
weight  of  the  body  M.  From  the  ends  of 
W  draw  the  line  P  at  right  angles  with 
the  plane  1.  Draw  F  parallel  with  the 
plane,  and  complete  the  parallelogram  as 
shown  by  Fig.  93. 

Then,  P  represents  the  force  with  which 
the  body  presses  against  the  inclined  plane,  and  F  the  force  which 
pulls  the  body  in  the  direction  of  the  plane. 

It  is  well  known  from  geometry  that  the  triangle  I,  b,  h  and 
W,  P,  -Fare  congruous. 

W:F=l:h    and    W:P=l:b. 

.f.™       and      P~E». 

Let  T  represent  the  time  in  seconds  from  the  instant  the  body  is 
let  loose  to  roll  without  friction  on  the  plane  until  it  has  attained  a 
velocity  V.  Then,  we  have 

F:M=  V:  T. 
Wh 


But 


and    M  =  — . 


11* 


Wh     W 

V:  T. 
hT     V 

,    VI 

I     '  9 
WhT     WV 

I            9 

V-lnritv       V     gkT 

1       9' 
and  time    2 

9*' 

126  ELEMENTS  OF  MECHANICS. 

Example.  The  proportional  length  of  the  plane  is  /=15  feet  in  a 
height  A  =  3  feet.  Required  the  velocity  of  the  body  after  having 
rolled  a  time  of  T=  6  seconds  from  rest  ? 

F=  32'17x3x6  =  38.6  feet  per  second. 
15 

G  =  acceleratrix,  with  which  the  velocity  of  the  body  is  accelerated 

on  the  inclined  plane. 
<7  =  32.17,  the  acceleratrix  when  the  body  falls  freely  or  vertically. 

Then     G:g=h:  I    and    £  =  ^. 

The  space  8  in  feet,  which  the  body  rolls  from  rest  in  the  time  T, 
will  be 


2          21          2       2G2gh 

Example.  How  far  will  the  body  roll  in  a  time  T=  25  seconds  on 
an  inclined  plane  of  height  A  =  15  feet  in  /  =  100? 


The  time  and  velocity  expressed  by  space  will  be 
08 

"" 


Example.  What  velocity  will  a  body  attain  in  rolling  a  space  of 
£=64  feet  on  an  inclined  plane  of  height  A  =  8  feet  in  £=100  feet? 


2x32.17x8x64     1Q1,,    ,  , 

P—*J =18.15  feet  per  second. 


The  velocity  which  a  body  attains  by  rolling  or  sliding  without  fric- 
tion or  other  resistance  on  an  inclined  plane  is  equal  to  that  attained 
when  falling  freely  a  height  equal  to  the  difference  of  level  rolled  on 
the  plane. 


WEIGHTS  ON  A  ROPE  OVER   A  PULLEY. 


127 


FORMULAS    FOR    BODIES   MOVING   FREELY   ON   INCLINED    PLANES 
UNDER  THE  ACTION  OF   GRAVITY. 


Velocity  in  feet  per  second. 

Space  in  feet  of  motion. 

Time  of  motion  in  seconds. 

-¥•  • 

.  1 

.  2 
.  3 

•\T1    7 

O                                                       R 

T-JP1              8 

F=VX2  QS. 

VTT* 

T                                  Q 

\2gh8 

\       I 

'  F' 

I  62.     FALLING   AND   RISING   OF   BODIES  CONNECTED   BY  A   ROPE 
OVER   A   PULLEY. 

The    accompanying    figure    represents    two  Fig.  94. 

equal  weights  IT  and  W,  hung  at  each  end  of  a 
rope  over  a  pulley.  As  the  weights  are  alike, 
the  one  balances  the  other,  and  there  will  be 
no  motion.  Hang  a  third  weight  w  under  the 
weight  W,  and  the  system  will  move  with  an 
accelerated  velocity  until  the  weight  w  reaches 
the  ground,  when  the  weight  W  and  W  will 
continue  to  move  with  a  uniform  velocity  equal 
to  that  attained  at  the  moment  the  weight  w 
stopped.  It  was  the  force  of  gravity  of  the 
weight  w  which  set  the  system  in  motion.  Sup- 
pose the  three  weights  to  be  united  into  one 
body,  and  rolled  or  dragged  on  a  level  plane, 
without  friction,  by  a  force  equal  to  the  force 
of  gravity  of  w,  that  body  would  move  with  the 
same  accelerated  velocity  in  equal  time  as  by  the  force  of  gravity 
when  hanging  on  the  rope. 

Let  M  denote  the  mass  of  the  three  weights  expressed  in  matts., 
and  F=  the  force  acting  to  move  the  system,  which  is  equal  to  the 
weight  w  in  pounds. 

T=  time  of  action  in  seconds. 

V=  velocity  in  feet  per  second  at  the  end  of  the  time  T. 


128  ELEMENTS  OF  MECHANICS. 


The  fundamental  formula  for  dynamics  of  matter  is 

M:F-T: 

V,        and  M  V=  F  T. 

F  T 
Jf=—  .     .       . 

1 

F  T 
V 

.    3 

v  =         •. 

V 

M 

MV 

0 

T^MV 

A 

F~     T   • 

A 

T~   F  '    '        ' 

.        rt 

Example  1.  A  mass  of  M  =•=  100  matts.,  which  is  3217  pounds,  is 
acted  upon  by  a  force  of  .F=  150  pounds  (not  necessarily  weight)  for 
a  time  of  T=8  seconds.  Required  the  velocity  T^when  the  force 
ceases  to  act  ? 

Formula  3.      V=  -  =  -  -  —  =  12  feet  per  second. 

Example  2.  What  force  F=  ?  is  required  to  give  a  mass  M=  360 
matts.  a  velocity  1^=20  feet  per  second,  the  time  of  action  being 
T=  15  seconds  ? 

„    MV    360x20 

Formula  2.     F=>  -  =  -  =  48  pounds. 
T  15 

Example  3.  What  time  T=  ?  is  required  to  give  a  mass  M  =  580 
matts.  a  velocity  F=36  feet  per  second,  the  acting  force  being 
F=  480  pounds? 


Formula  4.     T-          -  -  43.5  seconds. 

Example  4-  A  mass  of  unknown  quantities  is  acted  upon  by  a 
force  of  F*=  1680  pounds  for  a  time  of  T=  45  seconds,  in  which  it 
attained  a  velocity  of  V=  12  feet  per  second.  Required  the  quantity 
of  matter  in  the  mass  ? 


Formula  1.     If-          -  45  -  6300  matts., 

or  about  90  tons,  the  answer. 

Example  5.  A  mass  of  M=  84  matts.  has  a  velocity  of  148  feet 
per  second.  What  force  is  required  to  stop  that  mass  in  a  time  of 
T=  6  seconds  ? 

Formula  2.     F=~  ^=  84  X  14S  =  2072  pounds. 
T  6 

The  weight  of  a  body  in  pounds,  multiplied  by  0.0310849,  will  be 
the  mass  in  matts. 


FALLING  AND  RISING  OF  BODIES. 


129 


WThen  the  mass  is  expressed  by  ^ 
W-2-—.     ...    5 
P                                     fi 

weight  the  formulas  will  be 
v   9FT 

.    7 
.    8 

W 

gF'  '      ' 

£  63.     FALLING  AND  RISING  OF  BODIES  CONNECTED  BY  A  ROPE 
OVER  A  PULLEY. 

The  accompanying  figure  represents  two  weights  TFand  w  hang- 
ing on  a  rope  over  a  pulley.     The  heavy  weight  will  fall,  and  draw 
up  the  lighter  one  with  an  equal  velocity.     The  motive        pi   9g 
force  which  sets  the  system  in  motion  is  equal  to  the  dif- 
ference between  the  weights,  or  F  —  W—  w.     The  mass 
of  the  two  weights  is 


The  fundamental  formula  for  dynamics  of  matter,  as 
before  described,  is 

M:  F=T:  V. 

By  inserting  the  values  of  .Fand  M  in  this  analogy, 
we  have 


of  which  the  dynamic  momentums  are 


\ 


TT  i     -,. 
Velocity, 


Time, 


-rr 
F 


9  T(W- 
-  —  j 

Tr  +1O 


. 

g(W-w) 


Example.  The  large  TT=36  pounds  and  w  =  BQ  pounds.  What 
velocity  will  the  weights  attain  in  a  time  T=  3  seconds  from  the  time 
the  weights  start  to  move? 

,    32.17x3(36-30)     , 
F-  -  *  --  '  =  8.77  feet  per  second. 
36  +  30 


130  ELEMENTS  OF  MECHANICS. 

The  space  8,  which  the  weights  will  move  from  the  start  in  the 


Ume  T,  will  be  £=-        ........     6 

2 

The  formulas  will  hold  good  when  the  experiment  is  made  in 
vacuum,  but  when  the  weights  W  and  w  are  of  heavy  materials,  like 
that  of  metals,  the  difference  of  fall  in  air  and  in  vacuum  is  very  small 
and  can  be  neglected  for  ordinary  purposes.  The  weight  of  air  dis- 
placed by  solid  iron  is  only  0.00015  of  that  of  the  iron. 

When  the  weights  W  and  w  are  of  light  materials,  like  cork,  the 
difference  of  fall  in  air  and  in  vacuum  is  considerable.  The  weight 
of  air  displaced  by  cork  is  0.005  of  that  of  the  cork.  That  is  to  say, 
a  pound  of  cork  displaces  33  times  as  much  air  as  does  a  pound  of 
iron. 

A  piece  of  cork  which  we  may  suppose  to  weigh  1000  pounds 
Would  displace  5  pounds  of  air,  and  would  fall  much  slower  in  the  air 
than  in  vacuum  ;  but  let  the  cork  be  hung  on  the  rope,  Fig.  95,  in 
place  of  the  weight  W,  and  hang  another  weight  of  whatever  material 
weighing  5  pounds  at  w  ;  now  let  the  system  move  in  vacuum  under 
the  action  of  gravity,  and  the  weight  W  will  fall  with  the  same  ac- 
celeration of  velocity  as  when  it  falls  alone  in  air. 


264.     A   FORCE   ACTING   TO   MOVE   A    BODY   ON    A    HORIZONTAL 
PLANE   WITHOUT   FRICTION. 


The  body  M  is  supposed  to  be  at  rest  at  a,  and  a  force  F  is  applied 
to  move  it  from  a  to  b,  where  the  force  ceases  to  act.  If  the  time  of 
action  from  a  to  b  is  known,  then  the  velocity  V  at  b  is  found  by 
Formulas  3  and  7,  §  62.  After  the  force  J^has  ceased  to  act,  the  body 
will  continue  from  b  with  the  uniform  velocity  F  until  it  meets  with 
some  force  of  resistance,  say  at  c,  the  force  F'  acting  on  the  body  in 
an  opposite  direction  to  the  motion  until  it  stops. 

The  forces  F  and  F'  being  constant,  the  velocity  of  the  body  is 
uniformly  accelerated  from  a  to  b,  and  uniformly  retarded  from  c  to  d, 

for  which  the  mean  velocity  in  both  cases  will  be  =-  — . 


FORCE  ACTING   TO  MOVE  A  BODY.  131 

We  have  learned  (§  8)  that  space  S  is  the  product  of  time  and  ve- 
locity, 


in  which  Tis  the  time  from  a  to  b,  and  T'  that  from  c  to  d. 
S  =  space  from  a  to  5,  and 
s  =  that  from  c  to  d. 

VT  VT 

S  =  —     and    *--g-, 


,     ,.  ,  T_    2  £    2  a    . 

oi  which  K  =  -  =  —  =  -  = 


. 
T      T     M        M 

The  product  of  either  two  .of  these  equal  members  will  be  alike,  or 
2£   FT    2s    F'  T  28  FT    2  s  F'  T 


, 
T      M     T       M  TM          T  M  ' 

of  which         F  8=F'  s,     or     F  ':  F'  =s  :  & 

That  is  to  say,  the  force  F,  multiplied  by  its  space  of  action,  is 
equal  to  the  force  Ff,  multiplied  by  its  space  s,  and  the  products  are 
momentums  of  work,  which  will  be  explained  hereafter. 


3 


.! 


f-™.  .2 


Example  1.  A  force  F=  300  pounds  is  applied  on  the  body  M  from 
a  to  b,  a  space  of  8=24.  feet,  after  which  the  body  meets  with  a  re- 
sistance which  stops  it  in  a  space  of  s  =  0.5  feet.  Required  the  force 
of  resistance  F1  =  ? 

Formula  2.      F  =  —  =  ?°1^4  =  ^^^  pounds>  the  answer. 

The  forces  F  and  F  are  independent  of  the  weight  or  mass  of  the 
body  M,  for  the  reason  that  the  spaces  /S  and  s  are  inverse  as  the 
mass,  as  shown  in  the  proof  formulas. 

If  the  applied  force  F  is  equal  to  the  weight  of  the  body  M,  then 
the  body  will  move  on  the  horizontal  plane  with  an  equal  accelerated 
velocity,  as  if  it  was  falling  vertically  under  the  action  of  gravity ; 
therefore,  in  falling  bodies  the  acting  force  is  equal  to  the  weight  of 


132  ELEMENTS  OF  MECHANICS. 

the  body ;  which  is  the  same  as  to  say  that  the  weight  of  a  body  is  the 
force  of  attraction  between  the  earth  and  that  body. 

The  force  of  a  falling  body  is  equal  to  the  resistance  it  meets  with 
at  the  end  of  its  fall. 

Example  2.  A  steam-hammer  weighing  5  tons  falls  upon  a  red- 
hot  mass  of  iron,  in  which  it  sinks  s=1.5  inches  =  0.125  feet;  the 
height  of  fall  to  where  the  hammer  stopped  is  $=6.25  feet,  and  the 
force  F=b  tons.  Required  the  force  of  the  blow  of  the  hammer? 

Formula  1.     F  -  —  =  ^^  =  250  tons. 
a        0.125 


It  is  a  popular  expression  to  say  "  the  force  of  a  falling  or  moving 
body,"  or  "  the  force  of  the  blow  of  a  hammer,"  which  is  not  correct. 
There  is  no  more  force  in  a  moving  body  than  in  one  at  rest ;  in  fact, 
no  force  in  either  case.  The  force  of  inertia  overcomes  the  resist- 
ance which  the  moving  body  strikes,  and  which  resistance  is  the  force 
which  stops  the  motion  of  the  body. 


§65.     ON    DRIVING   A   NAIL  INTO  A   PIECE   OF   WOOD. 

Fig.  97. 


Fig.  97  represents  a  nail  driven  through 
a  piece  of  wood  by  a  weight  W  resting  on  the 
head  of  the  nail.  It  is  supposed  that  the  resist- 
ance to  the  nail  in  the  wood  is  equal  to  the 
weight  W,  so  that  the  slightest  additional  force 
would  cause  the  weight  to  drive  the  nail  down 
to  its  head,  as  shown  by  the  dotted  lines. 


In  driving  a  nail  into  a  piece  of  wood  the  resistance  is  not  uniform, 
for  the  deeper  the  nail  is  driven  in  the  greater  is  the  resistance ;  but 
the  mean  force  of  resistance  will  always  be  as  the  following  For- 
mula 2. 

Let  s  denote  the  space  which  the  nail  is  driven  into  the  wood 
by  the  weight. 


DRIVING  A  NAIL  INTO  WOOD. 


133 


Fig.  98. 


Let  the  same  nail  be  driven  into  the  same  piece  of  wood  by  the 
aid  of  a  lever,  as  represented  by  Fig. 
98.  The  force  F  acting  on  the  long 
lever  L,  presses  on  the  nail  equally  to 
the  weight  W.  The  force  of  resistance 
F'  to  the  nail  in  the  wood,  which  is 
equal  to  the  weight  W,  Fig.  97,  acts 
on  the  short  lever  /.  The  fulcrum  of 
the  lever  is  at  c. 

The   force  F  with  the  lever  L 
adjusted  so  that  it  balances  the  resist- 
ance F' ,  acting  on  the  lever  I. 


From  the  well-known  analogy  of  levers  we  have 


F:  F'  =  l:L. 

That  is  to  say,  the  weight  F,  Fig.  98,  is  so  much  smaller  than  the 
weight  W,  Fig.  97,  as  the  lever  I  is  smaller  than  L. 

Let  8  represent  the  space  which  the  weight  falls  in  pressing  down 
the  nail  in  the  wood,  and  5  =  the  space  the  nail  was  driven  in,  which 
is  the  same  as  the  space  s,  Fig.  97. 

It  is  well  known  in  geometry  that 

s:S-l:L, 

and  as  F :  F  =  I :  L, 

we  have  F :  F'  =  s  :  S,        and  F  S=  F'  s. 


F= 


8 
F8 


«-• 


F 

FS 


These  formulas  are  precisely  the  same  as  those  in  the  preceding 
paragraph ;  which  proves  that  if  the  weight  F  is  let  to  fall  from  an 
equal  height  S  directly  upon  the  head  of  the  nail,  the  latter  would 
be  driven  into  the  wood  precisely  the  same  distance  5  as  by  the  aid 
of  the  lever  and  weight  F,  Fig.  98. 

The  space  of  fall  S  must  include  the  space  s  of  penetration. 


134  ELEMENTS  OF  MECHANICS. 


§66.    PILE-DRIVER. 

The  formulas  for  pile-drivers  are  the  same  as  those  for  force  of  fall- 
ing bodies,  the  principles  of  action  being  the  same. 

NOTATION   OF   LETTERS. 

Fig.  99.  f7=  weight  of  the  ram  in  pounds. 

£=set  of  the  pile  in  feet  by  the  blow  of 
the  ram. 

R  =  resistance  in  pounds  to  the  pile  in  the 
ground. 

h  =  height  in  feet  which  the  ram  falls,  in- 
cluding the  set  8. 

V=  velocity  in  feet  per  second  with  which 
the  ram  strikes  the  pile. 

Works     Wh  =  RS.         .        .     1 

Weight  of  ram         W=^.    .        .     2 
h 

Height  of  fall  h  =  — .   .        .3 

Resistance  to  pile     R  = 


Set  of  pile 


. 


Striking  velocity        F=8v/A-&      .     6 

The  ram  is  raised,  either  by  hand  or  steam,  by  a  rope  over  a  pulley 
at  the  top  of  the  framing. 

In  using  the  above  formulas  some  allowance  should  be  made  for 
the  work  lost  by  the  ram  crushing  the  head  of  the  pile,  which  may 
amount  to  some  25  per  cent.,  or  use  only  75  per  cent,  of  the  actual 
weight  of  the  ram  for  TTin  the  formulas. 

Set    g-0-76*. 


STEAM-HAMMERS. 


135 


§  67.    STEAM-HAMMERS. 

The  illustration,  Fig.  100,  represents  a  steam-hammer  as  formerly 
made  by  Henry  G.  Morris  of  Philadelphia,  and  is  here  intended  only  for 
illustrating  the  dynamics  of  forging.  The  steam-hammers  now  made 
by  that  gentleman  are  differently  constructed. 

FORMULAS   FOR   STEAM-HAMMERS   FALLING   UNDER  THE   ACTION 
OF   GRAVITY. 


-rr   ,      ., 

\  elocity, 


Fig.  100. 


Work, 


Force 

Horse-power, 

pSn 


Time     T= 


NOTATION   OF 

W"  weight    of    the    ram    in 

pounds. 
8=  stroke  of  the  hammer  in 

feet. 
s  =  depth   of  penetration   by 

the  blow  in  feet. 
V=  velocity  in  feet  per  second 

of  the  blow. 
K=  work  in  foot-pounds  of  the 

blow. 

n  =  number  of  blows  per  min- 
ute. 
EP  =  horse-power    driving    the 

hammer. 


LETTERS. 

F=  mean  force  of  the  blow  in 

pounds. 
P  =  steam-pressure  in  pounds 

on  the  top  of  the  piston. 

when  such  is  used. 
T=  time  of  fall  of  the  hammer 

in  seconds. 
2  =  time  in  seconds   required 

to  raise  the  hammer,  the 

stroke  & 
p  =  steam-pressure  in  pounds 

on  the  under  side  of  the 

piston. 


136  ELEMENTS  OF  MECHANICS. 

Example.  The  weight  of  a  steam-hammer  is  W=  5000  pounds,  and 
is  lifted  $=4  feet;  the  penetration  of  the  blow  is  s  =  0.015  of  a  foot. 
Required  the  force  of  the  blow  ? 

-  1.338,333  pounds, 

or  600  tons,  nearly,  the  force  required. 

The  steam-pressure  p  under  the  piston  should  be 

= 


16  ** 

If  it  is  required  that  the  hammer  should  be  lifted  in  the  same  length 
of  time  as  that  in  which  it  falls,  then  make  p  =  2  W. 

Formulas  for  Steam-Hammers  falling  with  aid  of  Steam 
on  the  Top  of  the  Piston. 


Work  K=  (  W+  P)  (8+  «)  =  F  s. 

Force  fJ 


Horse-power 


Time 


16(TP+P) 

P=  Steam-pressure  on  the  tops  of  the  piston  in  pounds  (not  per 
square  inch). 

|68.    ON   FORCE   OF   MOVING   OR  FALLING   BODIES. 

This  subject  has  been  treated  under  different  heads  in  the  pre- 
ceding paragraphs — namely,  the  Formulas  2  and  6,  §  62,  give  the 
force  in  regard  to  velocity  and  time,  and  the  Formula  1,  §  65,  in  re- 
gard to  space.  It  now  remains  to  explain  the  subject  in  regard  to 
velocity  and  space. 

The  fundamental  analogy  of  the  four  elements  in  dynamics  of  mat- 
ter, as  before  repeated,  is 

M:  F=T:V, 

in  which  Fis  the  velocity  which  the  force  F  has  given  to  the  mass 
Min  the  time  T. 


FORCE  OF  MOVING  BODIES. 


137 


Space  8  is  the  product  of  velocity  V  and  time  T,  but  the  velocity 
in  the  space  8,  in  which  the  force  F  started  the  mass  M  from  rest  to 
the  velocity  F,  is  only  one-half  of  the  final  velocity  V,  or  the  uniform 
velocity  of  the  mass  after  the  force  ceased  to  act.  Therefore, 


8 


V  T 


of  which  r=— = 


28 


MV 
. 


M- 


MV* 

28' 

28  F 
V*   ' 


MV* 
2F' 

I2SF 


The  formulas  will  hold  equally  good  whether  the  force  Fseis  the 
mass  at  rest  to  motion,  or  brings  a  moving  mass  to  rest.  The  mass 
M  must  be  expressed  in  matts.  of  32.17  pounds  each. 

When  the  magnitude  of  the  mass  M  or  body  W  is  expressed  in 
pounds,  the  formulas  will  appear  as  follows : 


TFF2 


.    7 


.    8 


Example  1.  A  rifle-ball  of  W=  0.02  pounds  was  fired  with  a  ve- 
locity F=  600  feet  per  second  into  a  log  of  wood,  in  which  it  pene- 
trated $=0.9  of  a  foot.  Required  the  force  of  resistance  F<°*?  in  the 
wood? 


Formula  5.     F  •• 


WV1      0.02  x6002 


2x32.17x0.9 


=  124.34  pounds. 


Example  2.  A  projectile  of  W=  300  pounds  is  fired  from  a  gun 
S=  10  feet  long  with  a  muzzle  velocity  of  V—  800  feet  per  second. 
Required  the  mean  force  of  the  gunpowder  ? 


O  AA  v  O  AA2 

Formula5'   ^^il^To'298414^1"18- 


138 


ELEMENTS  OF  MECHANICS. 


§  69.     FORCE  OF  MOVING  BODIES  MEASURED  BY  A  SPRING. 

The  illustration  represents  a  body  B  moving  with  a  velocity  V 
between  two  equal  spiral  springs,  the  elasticity  of  which  has  been 
measured  by  experiments  that  when  the  spring  is  compressed  the 
space  s,  its  force  of  elasticity  is  f.  The  force  of  elasticity  of  spiral 
springs  generally  increases  uniformly;  that  is,  when  the  spring  is 
compressed  one-half  s,  the  force  will  be  one-half/,  and  so  on. 


The  height  of  the  dotted  triangle  may  represent  the  force  /  for  the 
corresponding  space  s,  from  which  we  see  that  the  mean  force  in  the 
space  s  is  equal  to  half  the  force  /. 

Let  S  denote  any  space  the  spring  may  be  compressed,  and  F'  = 
force  of  the  spring  corresponding  to  the  space  S.  Then 

S:8  =  I*:f,         8-—,        MdJ*-^. 
The  mean  force  in  the  space  8  will  be 


M  =  mass  of  the  body  B  expressed  in  matts. 
V=  velocity  of  B  in  feet  per  second. 


Mean  force,         F= 


M  V* 

28  ' 


F= 


, 

28  25 


.    4 


8=  V* 


f    MV* 
s'    8*    ' 


FORCE  OF  A  BODY  FALLING  UPON  A  SPRING. 


139 


Example  1.  The  mass  of  the  body  B  is  M  =  2  matts.  moving 
against  the  spring  with  a  velocity  F=  5  feet  per  second  ;  how  much 
will  it  compress  the  spring  ?  when  we  know  that  f=  50  pounds  will 
compress  it  s  =  0.2  feet  ? 

/O  2  x  2 

Formula  5.     8  =  5-J^-^  =  0.447  feet. 
\    50 

After  the  spring  is  compressed  by  the  blow  it  will  push  the  body 
back  with  the  same  velocity  to  the  other  spring,  where  the  same  oper- 
ation would  be  repeated,  and  so  the  body  would  move  between  the 
springs  for  ever  if  there  was  no  friction  or  other  resistance  to  retard 
and  finally  stop  the  motion.  It  is  not  necessary  that  the  springs 
should  be  alike,  only  that  they  should  be  perfectly  elastic  within  the 
limit  of  compression. 

Example  2.  A  body  ~B,  weighing  W=  50  pounds,  moves  against  a 
spring  with  a  velocity  F=25  feet  per  second,  and  compresses  it 
$=0.5  feet.  Required  the  mean  and  ultimate  forces  of  the  spring? 


Force, 


50x252 


2x32.17x0.5 


=  1000  pounds,  the  mean  force 


of  the  spring,  and  the  ultimate  force  will  be  double,  or  2000  pounds. 


\  70.     FORCE  OF  A  BODY  FALLING  FREELY  UPON  A  SPRING. 

The  weight  W  falling  the  space  8  will  compress 
the  spring  the  space  s.  After  the  weight  is  brought 
to  rest  the  spring  will  throw  it  up  again  to  the  same 
height  8,  and  so  the  weight  will  continue  to  ascend 
and  descend  for  ever  if  there  is  no  other  force  of  re- 
sistance to  retard  and  finally  stop  the  motion.  The 
weight  is  acted  upon  by  two  opposite  forces — namely, 
the  constant  force  of  gravity,  which  is  equal  to  the 
weight  of  the  falling  body,  acts  through  the  space  8, 
and  the  superior  force  of  the  spring  acts  in  the  oppo- 
site direction  in  the  space  5. 

Let  .F denote  the  mean  force  of  the  spring,  and/ 
that  at  the  greatest  compression  s.  Then  we  have, 
as  before  proved, 

W:  F=s:S,        and  WS=Fs. 
2W:f=s:S,      and2  TF£=/s. 


140 


ELEMENTS  OF  MECHANICS. 


The   greatest  force   of   compression  f=2F  for   ordinary   spiral 
springs. 


TF=  — 

p=W_8 
s 

a-**. 

S 

WS 
8  =   F  ' 


-./• 

28' 
.    2  WS 


2W 


.    5 


.    7 


In  these  formulas  the  spaces  S  and  s  can  be  expressed  by  any  unit 
of  length,  and  also  the  weight  W  and  force  F  by  any  unit  of  weight. 

Example  1.  The  weight  TF=4  pounds  falls  a  space  $=50  inches 
upon  a  spring,  which  it  compresses  s  =  3  inches.  Required  the  mean 
force  of  the  spring  ? 

Mean  force,        F  =  —  -  ^^  =  66.66  pounds. 
s  3 

Example  2.  It  is  known  by  experiments  that  the  spiral  spring  is 
compressed  s '  =  4.5  centimetres  by  a  weight  w  =  6.3  kilogrammes 
resting  upon  it;  from  what  height  S  must  a  weight  W=3.5  kilo- 
grammes fall  to  compress  the  spring  s  =  12.5  centimetres  ? 


Tins 


s  :  s'  =f:  w,  of  whichy= — — . 


insert  this  value  for  /  in  Formula  7,  which  will  give  the  required 
space. 


w  s' 


6.3x12.5* 

2  TTs'=2x3.5x4.5 


31.25  centimetres. 


\  71.     ELASTIC   BODIES  MOVING  AGAINST  HARD   AND   IMMOVABLE 
BODIES. 

We  have  now  considered  the  moving  body  to  be  perfectly  hard  and 
moved  against  an  elastic  spring,  but  the  body  may  also  be  perfectly 
elastic,  like  that  of  an  india-rubber  ball,  and  strike  against  a  perfectly 
hard  and  immovable  substance,  when  the  motion  of  the  body  will  be 


IMPACT  OR  COLLISION.  141 

the  same  as  when  the  non-elastic  body  moved  against  an  elastic 
spring. 

A  perfectly  elastic  body  B  falling  the  space  8  on  a  perfectly  hard 
and  solid  base,  will  be  flattened  by  the  blow,  or  the          Fig  W3 
product  of  the  weight  of  the  body  multiplied  by  its 
height  of  fall  3,  which  is  equal  to  the  force  F  of  elas- 
ticity multiplied  by  the  space  s  of  compression.     The 
force  of  elasticity  will  restore  the  primitive  form  of 
the  body  and  raise  it  to  the  same  height  S.    The  body 
will  so  continue  to  rise  and  fall  for  ever  if  there  is  no 
other  force  to  disturb  the  operation. 

If  the  falling  body  is  not  perfectly  elastic  it  will 
not  rise  to  the  same  height  from  which  it  fell,  but  the 
height  of  rise  is  a  measure  of  the  grade  of  elasticity 
of  the  body.  If  half  elastic,  it  will  rise  only  half  the 
space  S,  and  so  on.  The  elasticity  of  bodies  can  thus 
be  found  by  similar  experiments,  but  it  is  difficult  to 
find  a  perfectly  non-elastic  base  to  experiment  upon, 
because  all  bodies  in  nature  are  more  or  less  elastic.  A  cast-iron  ball 
falling  upon  a  massive  base  of  the  same  material  will  rebound  some, 
which  shows  that  cast-iron  is  partly  elastic,  and  even  a  cobble-stone 
falling  upon  a  solid  rock  will  indicate  some  elasticity. 


IMPACT    OR    COLLISION. 

g  72.    COLLISION  OF  PERFECTLY  HARD  AND  NON-ELASTIC  BODIES. 

A  moving  hard  body  striking  an  immovable  one  will  stop  at  the  col- 
lision. Practically,  bodies  are  not  perfectly  hard  nor  non-elastic,  and 
when  a  moving  body  stops  suddenly  against  an  immovable  one,  the 
momentum  is  destroyed  by  crushing  and  impressing  the  parts  in 
contact. 

A  moving  hard  body  striking  one  at  rest,  but  free  to  move,  will  set 
the  one  at  rest  in  motion,  so  that  both  bodies  will  move  with  equal 
velocity  after  the  collision. 

jjf  =  weight  of  a  body  of  velocity  V. 

m  =  weight  of  another  body  of  velocity  v. 

It  is  supposed  that  the  direction  of  motion  passes  through  the  cen- 
tres of  gravity  of  the  operating  bodies,  and  that  no  momentum  is  lost 
in  the  collision.  Although  the  bodies  are  denoted  by  M  and  m, 
which  means  mass,  they  can  be  expressed  by  any  unit  of  weight. 


142  ELEMENTS  OF  MECHANICS. 

The  sum  of  the  momentums  of  motion  of  the  bodies  will  be  alike 
before  and  after  the  collision,  and  when  the  bodies  are  perfectly  hard 
they  will  move  with  a  common  velocity  v'  after  contact. 


The  upper  sign  +  is  used  when  the  bodies  move  in  the  same  direc- 
tion, and  -  when  in  opposite  directions. 

273.    ONLY   ONE   OF   THE   HARD   BODIES   IN    MOTION. 

Fig.  lot  The  body  Jf  moves  with  a  velocity  V,  and 

strikes  the  body  m  at  rest,  or  v  =  o,  and  conse- 
/'    \%£f  quently  m  v  =  o. 


MV 


M+m 


M 


V-v1' 


Example  1.  A  hard  body  3f=  24  pounds  moves  with  a  velocity 
V=  42  feet  per  second,  and  strikes  the  body  m  =  16  pounds.  Re- 
quired the  common  velocity  of  the  bodies  after  collision  ? 

Velocity,          V  -  ^  —  =  25.2  feet  per  second. 

Example  2.  A  body  m  =  500  pounds,  at  rest  but  free  to  move,  is  to 
be  set  into  motion  of  v'  =  20  feet  per  second  by  a  body  M  =  80  pounds. 
Required  the  velocity  VI  of  the  body  M? 

Velocity,          V=  — =  145  feet  per  second,  the  answer. 

Example  3.  A  body  M=  250  pounds  moves  with  a  velocity  V=  64 
feet  per  second.  What  weight  m  ?  must  be  put  in  the  way  so  as  to 
reduce  the  velocity  of  Jf  from  64  to  5  feet  per  second? 

250(64-5)     , 
m ? '-  =  34oO  pounds. 

t) 

It  is  supposed  that  the  body  m  =  3450  pounds  is  free  to  move  after 
the  collision. 


IMPACT  OF  HARD  BODIES.  143 


874.     THE  TWO  HARD   BODIES  MOVE   IN   THE  SAME    DIRECTION. 

The  body  M  moves  fastest  until  it  reaches  Fi&- 105- 

m,  after  which  they  will  both  move  with  the 
common  velocity  v' . 

v'(M+  m) 


,     MV+mv 

v  = — — .        .         .     1 

M+m 

v_v'(M+m)-mv  g 

M 


Example  4-  The  body  M=  144  pounds  moves  with  a  velocity 
V—  72  feet  per  second,  in  =  480  pounds  arid  v  =  18.  Required  the 
common  velocity  v'?  after  contact  ? 

144x72+480x18     , 

Velocity.       t>'= —  =30.15  feet  per  second. 

J  144  +  480 


?  75.     THE   HARD    BODIES  MOVE  IN    OPPOSITE  DIRECTIONS. 

The  body  with  the  greatest  momentum  of  Fis- 106- 

motion  will  gain  on  the  other  body,  and  re- 
turn  it  with  the  velocity  v'. 


,     MV-mv 

.    I 
.    2 

m(v'  +v) 

M+  m 
v'(M+m)+mv 

V-v'    ' 
M(V-v') 

M 

m  — 

v  +v 

Example  5.  The  body  m  =  81  pounds,  and  moves  with  a  velocity 
y  =  36  feet  per  second  against  M=  27  pounds.  What  velocity  must 
the  body  3/have  to  stop  the  two  bodies  in  the  collision?  v'  =0. 

r    0(27 +  81) +  81x36 

Velocity,          F=  —      — L =  108  feet  per  second. 

27 


144 


ELEMENTS  OF  MECHANICS. 


|76.     COLLISION   OF   PERFECTLY   ELASTIC   BODIES. 

It  has  before  been  explained  that  when  a  perfectly  elastic  body 
strikes  an  immovable  body,  it  will  rebound  with  the  same  velocity  as 
that  with  which  it  struck  ;  but  if  it  strikes  a  movable  body,  some 
momentum  will  be  lost  by  the  striking  body  in  giving  motion  to  the 
one  at  rest ;  and  in  all  cases  the  sum  of  the  momentums  before  and 
after  collision  will  be  alike,  whether  either  or  both  bodies  are  non-, 
perfectly  or  partly  elastic. 

F=  velocity  of  JIf) 
v  =  velocity  of  in  ) 


877.  ONLY  ONE  OF  THE  ELASTIC  BODIES  MOVES  AND  STRIKES 
THE  ONE  AT  REST. 

Fig.  107. 


)=  V(M-m). 
v'(M+m)  = 


V 


V(M-m) 


2V  M 

*  M+m    ' 
t/(M-m) 

2M 
2V  M 


V-V  ' 

M(V-  V) 


M(v'-2  V) 


.  5 

.  6 

.  7 

.  8 


Example  6.  A  body  Jf=360  pounds,  moving  with  an  unknown 
velocity,  strikes  an  elastic  body  m  at  rest  and  of  unknown  weight. 
After  the  collision  it  is  found  that  the  velocity  of  M  is  V  =  42,  and 
that  of  m  is  t/  =  96  feet  per  second.  Required  the  weight  of  the 
body  ml 

.    0  360(96-2x42)      .. 

Formula  8.     m  = * =  90  pounds. 

96 


IMPACT  OF  ELASTIC  BODIES. 


145 


g78.     TWO   PERFECTLY   ELASTIC   BODIES    MOVING   IN   THE   SAME 
DIRECTION. 

The  body  J/ moves  fastest  until  it  reaches  the  body  m,  after  which 
it  will  move  slower  or  perhaps  return. 

Fig.  108. 


*%»+«)-  V(M— 

v'(M+m)  =  2M  V+v(m  -  M}. 


M+m 


M+m 
v'[V(M-m)  +  2 


2  MV+v(m-M) 
V'[2  M  V+v(m-M)] 
V(M-m)    ' 


,_m(V'  +  V-2v) 
V  -V 

=  M(2V-v-v') 
v'+v 


M- 


m-M 


Example  7.  The  body  m  =  800  pounds,  and  moves  with  a  velocity 
of  v  =  28  feet  per  second.  M=  360  pounds. 

It  is  required  to  give  such  a  velocity  to  the  body  M  that  when 
it  strikes  m  all  the  momentum  M  V  will  be  discharged  into  m ;  so 
that  M  will  be  at  rest  after  the  collision,  and  m  will  move  with  the 
two  momentums. 

mv'  =  MV+mv. 

Required  the  velocity  Fof  the  body  M?  and  v'  of  m? 

T.    0(360 +  800) -2x800x28 

Formula  7.      V=  — =  101.8 

360  -  800 

feet  per  second,  the  required  velocity  of  the  body  M. 

MV+mv  _8Mxl01.8  +  800x28 
m  800 

per  second,  the  velocity  of  m  after  the  collision. 
13  K 


146 


ELEMENTS  OF  MECHANICS. 


|79.     TWO   PERFECTLY   ELASTIC   BODIES   MOVE   IN   OPPOSITE 
DIRECTIONS. 

Fig.  109. 


V  = 


V(M-m)-2m  v 

M+m 
2MV-v(m-M) 

M+m 


v'[V(M-m-)-2mv] 
2MV-v(m-M)  " 
v>  _  V[2MV-v(m-]lf)] 
V(M-m)-2mv 


V'(M+m)  =  V(M-m)-2mv. 
2  M  V-  v(m  —  J/). 

1  M^m(2vv'+vfV-  V'v} 
=  2VV+  V'v  -  V'v' 
^M(2VV  +V'v-v'V 

2vv'+i/V-  V'v 
v=v(mV'-2mv'  -MV)   ? 


2MV'+mi/-Mv' 


m  V'-2mv'-MV 


.8 


Example  8.  M=  24  kilogrammes,  V=  16  metres  per  second,  in  =  10 
kilogrammes,  and  v  =  36  metres  per  second.  Required  the  velocities  V 
and  v'  after  the  collision? 

V  =  16(24-10)-2xlQx36  =  -  17.5  metres  per  second. 
24+10 

The  negative  sign  —  means  that  the  body  M  returns  with  that 
velocity. 

2x24x16-36(10-24) 

v'  =  -  *  --  '-  =  +29.88  metres  per  second. 
24  +  10 

The  positive  sign  +  means  that  the  body  m  continues  its  course 
with  a  velocity  reduced  from  36  to  29.88  metres  per  second. 


§  80.     COLLISION  OF   BODIES    OF   WHICH    EITHER    ONE    OR    BOTH 
ARE  ELASTIC,  NON-ELASTIC,  OR  PARTLY  ELASTIC. 

Let  E  denote  the  grade  of  elasticity  of  the  body  M,  and 

e  =  that  of  the  body  m. 

When  a  body  is  perfectly  elastic,  .Z^or  eis  equal  to  1,  and  when  per- 
fectly hard  or  non-elastic,  E  or  e  is  zero  or  0.  Therefore,  when  a  body 
is  half  elastic,  E  or  e  is  0.5,  and  so  on. 


IMPACT  OF  ELASTIC  BODIES. 


147 


2  81.     ONE  BODY  IN  MOTION  AND  ONE  AT  REST. 


v'(M+m)  = 

> 


M+m 
VM(l+e 


v'(M-Em) 
M(\+e) 

,  =  V  M(\  +  e) 
V  "    M-Em    ' 


Emv' 


Ev' 


M-Em  ' 


er 


.     5 


.    7 


.    8 


Example  9.  The  body  Jf  =  200  pounds,  and  elasticity  ^=0.75, 
struck  the  body  m  =  150  pounds,  and  elasticity  e  =  0.25.  After  the 
collision  it  was  found  that  the  velocity  of  m  was  v'  =  30  feet  per 
second.  Required  the  velocity  V  of  the  body  Ml 

T7,     30(200-0.75x160)    , 

Formula  3.      V  -  — V,,.,    ^^ — '  =  9.6  feet  per  second. 
200(1+0.25) 

g  82.     THE  PARTLY  ELASTIC  BODIES  MOVE  IN  ONE  DIRECTION. 

Fig.  111. 


M+m 


-e  M) 


M+ 


E)+v(m-eM) 


V(M-Em)  +  mr(\ 


v'v+V'e-V'V(I  +  E) 


V'v  +  v'VE-v'v(l  +  E) 

V'(M+m) 
M-Em  +  mv(l  +  E)' 

v'(M+m)-MV(l  +  E) 
m-eM 


148 


ELEMENTS  OF  MECHANICS. 


§83.     PARTLY    ELASTIC    BODIES   MOVING    IN    OPPOSITE    DIREC- 
TIONS. 

Fig.  112. 


V(M+m)  =  V(M-Em)-mv(l 
v'(M+m)  =  M  V(l  +  e)  —  v(m  —  e  J 

„     V(M-Em)-mv(l  +  E) 
M+m 

,     MV(l  +  E)-v(m-eM) 


Q 


E)-v(m-eM) 


f_  V'\M  V(l  +  E)  —  v(m- 
V(M-Em)-mv(l  + 


' 


7=- 


Vv-v'  E-v'v(l  +  E) 


V'(M+m) 


—  e  M 


•  Example  10.  A  perfectly  hard  body  M  =  125  pounds  is  moving 
with  a  velocity  of  V=  54  feet  per  second  against  a  perfectly  elastic 
body  w  =  480  pounds,  and  w  =  24  feet  per  second.  Required  the 
velocities  V  and  v'  after  the  collision  ? 

In  this  case  the  elasticity  of  M  is  E=Q,  and  that  of  m  is  e  =  1. 

Formula  1.      V  .54(125-0x480)-480x24(l  +  0)  _  ^  ^ 

per  second,  the  velocity  with  which  the  body  M  will  return. 

125x54(1  +  0)-  24(480-  1x125)         _n__ 
Formula2.    */=—        2  -  -^  =-  2.92  feet  per 


second,  the  returning  velocity  of  the  body  m. 

Both  the  bodies  will  return  in  opposite  directions  after  the  collision. 


DYNAMICS  OF  MATTER.  149 

DYNAMICS    OF   MATTER. 

§84.     ON   POWER   AND   WORK   APPLIED   TO   MATTER. 

The  three  physical  elements,  force,  motion  and  time,  with  their  com- 
bination in  space,  power  and  work,  have  already  been  explained — 
namely, 


Elements. 

Force    =F  -i -  -  '.  .     1 

Motion  =  V  ' .      ".'""  .    2 

Time     =T    .  .3 


Functions. 

8  =  V  T  .  -.  •  4 

Power,    P-FV  .  .5 

Work,     K=FVT  .    6 


In  the  application  of  this  law  upon  matter,  a  fourth  element,  mass, 
enters  into  the  combination  which  constitutes  the  dynamics  of  matter. 

The  relation  between  these  four  elements  has  already  been  ex- 
plained in  regard  to  force  and  motion,  and  it  now  remains  to  treat 
them  in  regard  to  power  and  work. 

The  fundamental  analogy  in  dynamics  of  matter  is 


Multiply  both  the  momentums  by  the  velocity  V,  and  we  have  the 
work  K=  M  V*  =  F  V  T,  which  means  that  the  work  consumed  in 
giving  the  mass  M  the  velocity  V  is  F  V  T,  or  the  product  of  force, 
velocity  and  time. 

In  the  formula  for  work,  K=  F  V  T,  V  means  the  mean  velocity 
of  the  force  Fin  the  time  T. 

When  a  constant  force  sets  a  body  (at  rest,  but  free  to  move)  in 
motion,  the  velocity  at  the  start  is  zero  or  0,  but  is  accelerated  to  V 
in  the  time  T]  and  the  mean  velocity  in  that  time  is  only  one-half  of 
the  final  velocity;  therefore,  if  F  means  the  final  or  uniform  velocity 
of  a  moving  body  after  the  force  has  ceased  to  act  upon  it,  the  work 
stored  in  the  body  will  be 


The  space  in  which  the  body  was  set  in  motion  is 

<&'-*  VT, 
which,  inserted  in  the  formula  for  work,  will  be 


A  body  lifted  vertically  a  space  or  height  S  against  the  force  of 
13* 


150  ELEMENTS  OF  MECHANICS. 

gravity,  but  without  regard  to  velocity  and  time,  the  work  accom- 
plished is  K=FS,  but  the  force  F  is  equal  to  the  weight  W  of  the 
body,  and  K=  W  8. 

Now,  let  the  body  fall  freely  to  the  same  spot  from  which  it  was 
lifted,  and  the  force  of  gravity  will  perform  an  equal  work, 

K=  WS. 

In  the  falling  work  the  velocity  and  time  might  be  entirely  differ- 
ent from  that  in  the  lifting  work,  but  their  product  will  be  equal  in 
both  cases ;  that  is,  the  time  multiplied  by  the  velocity  of  the  lifting 
work  is  equal  to  the  time  multiplied  by  the  mean  velocity  of  the  fall- 
ing work. 

The  space  $= \  V  T,  in  which  V  means  the  velocity  the  falling  body 
has  attained  when  striking  the  ground:,  and  Tthe  time  of  fall.  Insert 
this  value  of  space  in  the  formulas  for  work,  and 

J5T-  W    V  T. 

The  time  of  fall  is 


"Work  K  =  M\  V*,  which  was  to  be  proved. 

That  is  to  say,  the  work  stored  in  a  moving  body  is  equal  to  the 
mass  multiplied  by  half  the  square  of  its  velocity. 

$85.     DYNAMICS   OF   MATTER   REPRESENTED   GEOMETRICALLY. 

The  combination  of  the  physical  elements  and  functions  with  matter 
can  be  represented  by  the  corresponding  geometrical  quantities,  as 
explained  in  §  2,  and  illustrated  by  Fig.  113. 

Let  the  arrow  F  represent  the  magnitude  and  direction  of  a  con- 
stant force  applied  on  a  body  M at  rest,  but  free  to  move.  The  body 
M  will  then  be  started  and  moved  with  an  accelerated  velocity  as 
long  as  the  force  acts.  Let  the  line  T  represent  the  time  of  action 
of  the  force,  in  which  the  body  moves  from  a  to  b,  and  let  the  vertical 
line  F  represent  the  velocity  attained  at  b.  The -force  F  ceases  to 
act  at  b,  and  the  body  will  continue  its  motion  with  the  uniform 
velocity  V  until  some  other  force  is  applied  to  change  and  stop  its 


DYNAMICS  OF  MATTER. 


151 


motion.  At  b'  the  body  meets  a  force  F  in  opposite  direction  to  its 
motion  which  will  be  stopped  at  a'  in  the  time  T .  Whilst  the  body 
moved  from  b  to  b',  the  space  S=  VT",  its  momentum  of  motion  was 

M  V=  F  T=  FT,    or    F :  F  =  T  :  T. 
The  illustration  is  so  proportioned  that  F  =  2  ^and  T=2T'. 

Fig.  113.  d> 


Draw  the  line  a  k,  and  the  area  of  the  triangle  a,  £,  &  represents 


VT 


which  the  force  F  passed  through  in  the  time  T. 

Draw  from  a  the  line  a  e  at  any  convenient  angle  to  represent  the 
force  F,  and  complete  the  wedge  or  prism  a,  b,  c,  d,  e.  Then  the 
cubic  contents  of  that  prism  will  represent  the  work  K  consumed  in 
giving  the  mass  M  the  velocity  V. 

T 


Complete  the  same  operation  at  the  other  end  of  the  illustration, 
and  the  prism  a',  b',  </,  d',  J  will  represent  the  work  consumed  in 
bringing  the  mass  MtQ  rest. 

The  cubic  contents  of  the  two  prisms  are  alike,  or  the  works 

F  V  T     F  V  T     M  V2 

=      2 2  ~~2~ ' 

2JT=  F  V  T=  F  V  T  =  MV\ 

Work  is  always  K=  F  V  T,  in  which  the  velocity  V  means  the 
mean  velocity  in  the  time  T,  but  in  the  case  of  a  moving  body  the 
velocity  V  is  double  the  mean  velocity  of  the  force  which  put  the 
body  in  motion  or  brought  it  to  rest.  Therefore  the  work  of  a  mov- 


152  ELEMENTS  OF  MECHANICS. 

FVT 

ing  body  is  expressed  by  K=  -  ,  when  V  means  the  uniform 

velocity  of  the  body. 

The  area  of  the  rectangle  a,  b,  c,  e,  or  the  base  of  the  prism,  repre- 
sents the  momentum  of  time  F  T,  which  is  equal  to  the  momentum 
of  motion  of  the  moving  mass  M  V. 

The  power  in  operation  at  any  time  i  is  represented  by  the  area 
of  the  rectangle  f,  g,  k,  i,  where  the  velocity  of  the  mass  or  force  is 
v,  or  the  section  Fv  of  the  prism  is  the  power  at  the  time  t.  The 
power  varies  with  the  velocity  as  long  as  the  force  acts,  after  which 
there  is  no  power  in  operation.  The  mean  power  required  to  set  in 
motion  or  bring  to  rest  a  body  is 


2  T 


The  work  expended  in  setting  a  body  in  motion  is  restored  or  re- 
utilized  in  bringing  the  moving  body  to  rest ;  but  the  elements  of  the 
expended  and  utilized  works  need  not  be  alike. 

Either  one  or  two  of  the  three  physical  elements  can  vary  ad  lib- 
itum, but  only  at  the  expense  of  the  remaining  two  or  one  of  them,  so 
that  the  product  of  the  three  elements  which  constitute  the  work 
is  alike  in  both  cases. 

The  two  works  may  be  distinguished  as  follows : 

Primitive  work,  F  V  T=  F'  V  Tf,  the  realized  work. 

The  work  expended  in  setting  the  body  in  motion  is  the  primitive 
work,  and  that  restored  in  bringing  the  body  from  motion  to  rest  is 
the  realized  work. 

The  work  of  steam  in  a  steam-engine  is  the  primitive  work,  which 
is  equal  to  the  realized  work  executed  by  the  engine,  less  that  con- 
sumed by  friction. 

The  two  works  can  be  distinguished  in  different  ways — as  positive 
and  negative ;  action  and  reaction ;  primary  and  secondary ;  original 
and  final;  cause  and  result,  etc. 

If  we  call  the  decomposition  of  carbonic  acid,  forming  vegetation, 
a  primary  work,  then  the  burning  of  the  carbon,  forming  carbonic 
acid  or  generating  heat,  will  be  a  secondary  work.  When  the  gene- 
ration of  heat  is  considered  a  primary  work,  the  generation  of  steam 
may  be  termed  the  secondary,  and  so  on. 


DYNAMICS  OF  MATTER.  153 

Example  1.  The  mass  M  =  12  matts.  is  set  in  motion  by  a  force 
.F=36  pounds,  acting  for  a  time  of  jT=6  seconds.  Required  the 
work  consumed  on  the  mass,  and  the  velocity  at  the  end  of  six 
seconds?  also  the  space  S  in  which  the  force  acts,  and  the  mean 
power  of  the  operation  ? 

The  velocity      F=  —  =  ^^-  =  18  feet  per  second. 
AT         12 


. 
=  1944  foot-pounds. 


The  work  k  will  be 

.    MV    12x18* 
K=  -  =  - 

2  2 

This  work  ought  to  be  equal  to  %F  V  T,  or 

K=  -  =  -  -  -  =  1944  foot-pounds,  the  same  result. 
2  2 

The  space  in  which  the  force  acted  in  the  six  seconds  will  be 

VT     18x6 
~2~  =  ~2~ 

The  mean  power  in  operation  will  be 
p_FV_  36x18 


Example  2.  After  the  mass  M=  12  matts.  received  the  uniform 
velocity  of  V-  18  feet  per  second,  it  encountered  a  force  of  resistance 
F'  =  72  pounds.  In  what  time  can  that  force  stop  the  body  ? 

,    MV    12x18 
^==          -  =  3  seconds. 


The  work  consumed  in  stopping  the  body  will  be 
FVT    72x18x3 


which  is  the  same  as  the  work  which  set  the  body  in  motion. 


154  ELEMENTS  OF  MECHANICS. 

CALCULUS, 

APPLIED   TO   DYNAMICS. 

§  86.     Space. 

Fig.  114. 

The   function    space    is    generated    by  the   two 
elements  velocity  and  time. 

8=  VT. 

Let  the  function  linear  space  be  represented  by  the  area  of  a  rec- 
tangle of  which  the  velocity  V  represents  one  side,  and  the  time  I7  the 
other.  Assume  the  velocity  Fto  be  constant,  and  the  space  s  to  vary 
with  the  time  t,  then  we  have  the  differentials, 

8»  =  V^,    of  which  the  velocity    F=  — . 

6t 

fis 

This  expression  —  can  be   inserted   for  the   velocity  V  in   any 
ct 

dynamical  formula ;  for  instance, 

F-.M-V-.T,    of  which    F--^. 

Insert  — •  for  V  in  this  formula,  and  we  have 

ct 


of  which  the  space     8= 


2 

FT1 
2M' 


That  is  to  say,  a  force  F  acting  on  a  mass  M  free  to  move  will  pass 
through  the  space  Sin  the  time  T,  according  to  the  formula. 

Fig.  us.  We  may  also  consider  the  time  T  to  be  constant 

and  the  velocity  variable,  as  represented  by  Fig.  115. 


S=VT,    and    QS=  T8v,    of  which    T-— . 


DYNAMICS  BY  THE  CALCULUS.  155 


This  expression  —  can  be  inserted  for  the  time  T  in  any  dynam- 

cv 

ical  formula. 

.    MV 


,  o,  ,*-*?., 

which  is  the  formula  for  work  in  dynamics  of  matter.     The  work  re- 
quired to  bring  a  mass  from  rest  to  motion,  or  from  motion  to  rest,  is 


§  87.     Power. 

The  function  power  is  the  product  of  force  and 
velocity,  which  can  be  represented  by  a  rectangle. 

Power         P=FV   and    tiP=V8f, 

r\  TTJ 

when     Fis  constant,    or     V=  — . 

This  expression can  be  inserted  for  the  velocity  V  in   any 

If 
dynamical  formula ;  for  instance, 

Work         JT= 


K=PT; 

which  is  to  say  that  work  is  the  product  of  power  and  time. 

We  may  also  assume  the  force  to  be  constant  and  Fig.  in. 

the  velocity  variable,  as  represented  by  Fig.  117. 

Power        P-FV,    and    8P-F8v,  V 


of  which        F=—.  . 


P 


This  expression  --  can  be  inserted  for  the  force  -Z^in  any  dynam- 

cv 

ical  formula  ;  for  instance, 

FS    S8P 


C 


Power 

T8v  =s  »P,  •  or    S  =  V  T. 


156 


ELEMENTS  OF  MECHANICS. 


§8.    Work. 

The  function  work  is  represented  by  the  cubic  contents  of  the  par- 
allelopipedon  F  V  T^  of  which  the  side  P"  represents  force,  F  velocity 
and  T  time. 

R«- "«•  The  area  of  the  rectangle  F  V  represents 

power  and  V  T  space. 

Work,         K=FVT=PT. 

Assume  the  power  P  to  be  constant  and 
the  time  T  variable,  we  have 

-7         Dfi,  A     T>       %K 

CK  =  Jr  ct.          ana  Jr  = . 

8t 

This  expression  -— -  can  be  inserted  for  the  power  P  in  any  dynam- 

ct 

ical  formula. 

Fig.  119  represents  S  =  V  T  to  be  constant,  and  the  force  F  var- 
iable. 

Fig.  119. 


#=  -  — ,  which  can  be  inserted  for  space  in 
any  dynamical  formula. 

Work,  K=FS,    or    8 


of  which 


. 

8s 


f\  jf 


-^-  ,  can  be  inserted  for  force  in  any  dynamical  formula. 

C'S 

Either  or  all  the  elements  F,  V,  T  may  be  variable  in  the  execu- 
tion of  work. 


3  89.    POWER  AND  WORK  WITH  CONSTANT  FORCE  AND  VARIABLE 
VELOCITY. 

Suppose  a  constant  force  F  to  be  applied  on  a  mass  M  free  to 
move,  Fig.  120,  it  will  set  the  mass  in  motion  with  an  accelerated 
velocity.  When  the  force  has  acted  for  a  time  t,  the  velocity  will  be 
v  and  the  power  P=  Fv. 


F:  Jf=  v:t,  of  which  z!-— , 
•  F 


and  tit  = 


M&v 


CONSTANT  AND    VARIABLE  FORCES. 


157 


The  differential  work  M-P&- 
FvMVv 


/•  MV* 

Work  .fiT=  I  Jf  v  8v  =  -  ,  which  is  the  well-known  formula  for 


work  in  a  moving  body. 


Fig.  120. 


The  force  F  ceases  to  act  at  the  time  T,  and  the  mass  M  will  con- 
tinue to  move  with  the  uniform  velocity  V  and  generate  the  space 
S  =  V  T".  There  is  no  work  accomplished  in  the  time  T" ',  but  the 
mass  will  continue  to  move  until  some  force  F'  is  applied  in  opposite 
direction  to  retard  and  finally  stop  the  body  in  the  time  T' . 

The  work  which  stopped  the  body  is  equal  to  that  which  set  it  in 
motion,  and  which  is  represented  by  the  volume  of  the  two  prisms, 

= ,    or    FT- FT'. 

The  two  momentums  of  time  F  T  and  F'  T'  are  alike,  and  each 
equal  to  the  momentum  of  motion  M  V. 

I  90.     FORCE   VARIABLE   WITH   THE   TIME   OF   ACTION. 

Fig.  121. 

When  a  force  F,  acting  on 
a  mass  M,  increases  as  the 
time  of  action,  so  that  F=  C  T, 
C  being  a  constant  factor, 
the  relation  between  time,  ve- 
locity and  space  will  be  as 
follows : 

F-M=  V-  T 

•2'      -      -     —    cm' 


158                            ELEMENTS  OF  MECHANICS. 

•*-*% 

Ce&  CM^S      **   MS 

J           J     C  '         3  "    C  ' 

Time 

T_JzWs 

1 

Space 

*-fj£  

2 

Velocity 

y_F_TjCT 

K/T        9    TL-f'           •           •           • 

jXL            —    J>1_ 

3 

The  velocity  is  as  the  square  of  the  time,  and  the  curve  is  therefore 
a  parabola  tangenting  the  time  with  its  vertex  at  the  start  of  motion. 

Example.  Suppose  it  be  known  that  the  force  f=  32  pounds  after 
having  acted  on  the  mass  Jf=50  matts.,  for  a  time  of  2  =  4  seconds. 
Required  the  velocity  and  space  when  the  force  has  acted  for  a  time 
of  T=  36  seconds  ? 

f-Ct,    and    C=£  =  —  =  8. 


Velocity  F=  ^—^-  =  103.7  feet  per  second. 

2  x  50 


Space  8-  ^-^-  =  2488.3  feet. 

3x50 

The  area  bounded  within  V  ^represents  the  space  passed  through  ; 
the  area  of  the  section  /  1  represents  the  power  in  operation.  The 
area  of  the  base  triangle  represents  the  momentum  of  time  ^  F  T= 
M  V,  the  momentum  of  motion,  and  the  cubic  content  of  the  figure 

M  V* 
represents  the  work  done  by  the  force  F,  which  is  K=  -  . 


The  force  /=  C  t. 

t*    Mv 


Time 


VARIABLE  FORCES. 


159 


The  same  time  is  obtained  from  Formula  2. 

When  the  mass  is  expressed  by  weight  W,  the  formulas  will  be 


Time 


Velocity 


Space 


Time 


I3W8 


§91.     FORCE   VARIABLE   INVERSELY   AS   THE  TIME   OF   ACTION. 

A  force  F  is  applied  to  move  a  mass  M  toward  o,  and  the  force 
diminishes  as  the  time  of  action ;  so  that  in  a  time  T  the  force  /  is 
reduced  to  o.  Force/-  C( T-  £). 

Fig.  122. 


Velocity 


Space 


160 


ELEMENTS  OF  MECHANICS. 


When  the  formulas  are  integrated  for  the  whole  time  Tt  we 

ri  rpt 

Velocity 


F= 


3 


Time 


8- 


BSJf 

\-c~ 4 

CT* 

'3  M' 

These  formulas  are  the  same  as  those  in  the  preceding  paragraph. 
When  the  body  M  passes  the  centre  o,  the  force  /  becomes  negative 
and  stops  the  motion  at  a ;  which  operation  is  accomplished  in  equal 
length  of  time  T  and  space  /S  as  that  in  which  it  was  set  in  motion 
from  rest  to  the  velocity  Fat  o. 

Example.  A  force  F=  240  pounds  sets  a  body  M=  15  matts.  in 
motion,  so  that  in  a  time  T=  30  seconds  of  action  the  force  is  reduced 
to  0.  Required  the  velocity  of  the  body  when  arriving  at  0,  and 
what  space  it  has  passed  through  ? 


Formula  3.         Velocity 
Formula  5.         Space 


J         /"Y  ^iV          o« 

and    G  =  —  = —  80. 

T      30 

F=  ^p  =  2400  feet  per  second. 


3x15 


The  body  arrived  at  0  with  a  velocity  of  2400  feet  per  second,  and 
it  will  continue  to  move  with  that  velocity  until  some  force  is  applied 
to  change  it;  but  if  the  force/  continue  to  act  negatively  in  the  same 
ratio,  the  body  will  be  brought  to  rest  at  a,  1600  feet  from  0. 

§92.    FORCE   VARIABLE   AS  THE  SPACE   OF  ACTION. 

Assume  F=  CS,  in  which  G is  a  constant  factor. 
Fig.  123.  F:  M  =V:  T. 


MV^MV^ 

F       CS      Csct 


MVs 
Cs  ' 


/'*-/ 


MS* 


rjn 

— 


VARIABLE  FORCES.  161 


Time  T-J^¥  hyp.kg.8.  .    1 

FT     C8T     CS8S 

Velocity      F=  — —  = 

M        M 


Velocity      F-tf-r.     .  :'  -    2 

•    •    •'•    •  s 


Example.  The  force  F=  3  pounds  at  a  distance  s  =  1  foot,  and  the 
mass  M  =  8  matts.  Required  the  time  and  velocity  of  the  body  at  a 
distance  #=16  feet? 


Time,  T=  \F        hyp-^0- 16  -  S-845  seconds. 

\    3 

Velocity,      F=  16-J-  =  9.8  feet  per  second. 
When  the  mass  is  expressed  by  weight  W,  we  have 


Time,  T=hyp.kg.S.          ,        .4 


Velocity,    F=^^p          .        .        .        .5 

I&J^-^JJP    '.    .    .  '.;• 


14* 


162 


ELEMENTS  OF  MECHANICS. 


\  93.     FORCE  VARIABLE  INVERSELY  AS  THE  SPACE   OF  ACTION. 

A  force  F  is  applied  to  move  a  mass  M  toward  o  and  diminishes  a 
the  space  S,  so  that  at  o  the  force  is  reduced  to  nothing. 
Force /-<7  08V  «). 


Fig.  124. 


F:  M  =  V:  T, 


Ftit  _  C(S-s) 
M  "        M 

C 


=  —  ,        then  8»~~(£-  «)8s. 


Velocity,       » 


When  the  velocity  is  integrated  through  the  whole  space  S,  we 
have  *  =  S  and  the 


Velocity,          V'S\^ 2 

The  time  of  action  through  the  space  $  will  be 


Time,         T=^~ 3 


That  is  to  say,  the  time  required  to  move  the  body  to  o  is  inde- 
pendent of  the  distance  S  as  long  as  F=  C  8. 


VARIABLE  FORGES. 


163 


Example.  A  mass  M=  8  matts.  is  acted  upon  by  a  force  F*=  48 
pounds  at  a  distance  $=16  feet.  Required  the  time  and  velocity 
atO? 

~  S  "  16  ~ ' 

Velocity,    V=  16-J-  -  9.8  feet  per  second. 

\8 

Time,  T=  -J-  =  1.633  seconds. 

\3 

When  the  mass  is  expressed  by  weight  W,  we  have 
Velocity,          V 


Time, 


T-     I 
*    * 


After  the  body  has  passed  the  centre  o  the  force  F  is  negative  and 
stops  the  body  at  an  equal  distance  -  8, 


£94.     FORCE  VARIABLE  INVERSELY  AS  THE  SQUARE  OF  THE  DIS- 
TANCE FROM  THE  FORCE  OF  ACTION  TO  A  GIVEN  POINT. 

The  force  of  attraction  between  bodies  is  inversely  as  the  square 
of  their  distance  apart. 

Let  M  and  m  denote  the  n*- 125- 


in  matts,  and  S  their  distance 
apart  in  feet.  The  force  of 
attraction  between  them  is 


Mm 


18 


SP  =  28693080,   the   coefficient 
of  attraction,  §  51. 

Suppose  the  mass  M  to  re- 
main stationary  and  draw 
the  mass  m  to  it.  s  =  distance  moved  by  m. 


F: 


Mm 


164 


ELEMENTS  OF  MECHANICS. 


w    '  8T 


Time,         I 

This  is  the  time  in  which  the  mass  m  will  move  a  space  8. 

Mm  T          M  T 


vi    •  rr 

M 


--t    then     F- 
'     »F'  9(8-  8? 


F8F- 


Velocity 


I    2M 


That  is  to  say,  the  velocity  is  inversely  as  the  square  root  of  the 
distance  between  the  attracting  bodies. 

|  95.    RESISTANCE    OF   AIR   TO  A    MOVING   BODY. 

A  mass  M  rolling  without  friction  on  a  level  plane  a  b  arrives  with 
a  velocity  V  at  a,  where  it  is  left  to  work  its  own  way  against  the  re- 
sistance of  the  air. 

The  force  of  resistance  to  a  body 
moving  in  any  perfect  fluid  is  as 
the  square  of  the  velocity ;  where- 
:-  fore  the  resistance  at  a  can  be  rep- 
resented by  F°  C  V*,  of  which  C 
is  a  constant  to  be  determined 
hereafter. 

tt  3  When  the  body  has  moved  from 

a  to  b  in  the  time  t  the  resistance  of  the  air  has  reduced  the  velocity 


Fig.  126. 


RESISTANCE  OF  AIR.  165 

from  Fto  V ,  or  V  =  V-v.    The  force  of  resistance  of  the  air  at  b 
can  therefore  be  represented  by/=  C(  V-v)2. 

F:M=V:  T,    and    T=—. 
F 


Sl  f 


r 

J  < 


<7(F-t;)2     o(F-t>) 

This  formula  gives  the  time  in  which  the  velocity  V  is  reduced  to 
F'=(F-v). 

The  body's  motion  will  be  stopped  when  V  =  0  or  whenv  =  F,  which 
should  be  when 

M  M  M 

•  =  00  — 


q  F-  F')     C7  F  C  F 

That  is  to  say,  the  resistance  of  the  air  requires  an  infinite  length 
of  time  to  stop  the  body,  or,  more  correctly,  the  body  will  never  be 
stopped  by  th*at  means  alone,  but  the  velocity  will  be  reduced  so  that 
no  motion  conld  be  perceptible  in  days  or  years. 

By  solving  the  Formula  1  we  find  the  velocity 


M+CtV' 


Call  /8r=the  linear  space  the  body  would  move  through  with  the 
velocity  Fin  the  time  t  without  the  resistance  of  air,  or  $=  V  t. 
8'  =the  actual  distance  moved  in  the  time  t  against  the  resistance 

of  the  air. 

s  =  the  retarded  distance  in  the  time  t,  or  s  =  /S—/S'. 
From  the  Formula  1  we  obtain 


CVtlt 

M+Ct' 
V 


r_  Vt-—\hyp.log.(M+CV£)-hyp.l<)g.M\. 
C  [ 


166  ELEMENTS  OF  MECHANICS. 

This  linear  space  is  represented  by  the  area  of  the  figure  bounded 
within  the  lines  c  tv. 

The  space  S'  which  the  body  actually  moves  in  the  time  t  will  be 


-  TT| 

c  L 


C  Vt)  - 


This  linear  space  is  represented  by  the  area  of  the  figure  bounded 
within  the  lines  V  d  V  c. 

The  linear  space  S  which  the  body  would  have  moved  through 
with  the  constant  velocity  Fin  the  time  Tis  represented  by  the  area 
of  the  rectangle  Vt  in  the  figure. 

The  area  of  the  base  bounded  within  the  lines  Ft  e  represents  the 
momentum  of  the  resistance,  which  is  equal  to  the  momentum  M  v. 

The  volume  of  the  figure  bounded  within  the  lines  fv  t  represents 
the  work  done  by  the  resistance  of  the  aif,  and  which  is  equal  to  the 
work  |  J/V. 

The  time  required  for  the  body  to  move  through  the  space  S'  is 
found  as  follows  ; 

Call          X=M+CVt.       .        .     .  .        .6 


.     7 


8 


The  common  logarithm  multiplied  by  2.30258509  is  the  hyperbolic 
logarithm. 

When  the  time  t  occupied  by  the  body  in  passing  through  the  space 
S'  is  correctly  known,  the  initial  velocity  I7"  will  be 


at 


§  96.     Coefficient  C. 

It  now  remains  to  find  the  coefficient  C  for  the  resistance  of  the  air. 

The  resistance  is  equal  to  the  weight  of  a  column  of  air  with  a  base 
equal  to  the  projecting  area  of  the  moving  body,  and  a  height  equal 
to  that  from  which  a  body  falls  and  attains  the  same  velocity  as  that 
of  the  resistance. 

A  cubic  foot  of  air  of  temperature  60°  Fahr.  and  under  a  pressure 
of  30  inches  of  mercury  may  be  assumed  to  weigh  530  grains. 


DYNAMICS  Of  MATTER.  167 

A  =  area  of  resistance  in  square  feet  of  the  moving  body. 
h  =  height  in  feet  of  the  column  of  air. 
The  force/  of  the  resistance  of  the  air  will  then  be  in  pounds. 

,530,4  A 


The  height    A  = 


7000 
F2 


.  10 


. 

7000x20 
Of  which  the  coefficient     a**--.    .        .11 


When  the  area  is  expressed  in  square  inches  a,  we  have  the  co- 

efficient 


122367 


12 


A  or  a  means  a  flat  surface  at  rig"ht  angle  to  the  direction  of  motion. 
For  a  cylinder  moving  with  its  convex  side  to  the  motion  the  area  of 
resistance  is  one-half  of  the  projecting  area.  When  the  moving  body 
is  spherical,  the  area  of  resistance  is  one-quarter  of  the  projecting  area. 

D  =  diameter  in  feet,  and  d=  diameter  in  inches  of  a  moving  sphere. 

^4=  0.19635  D*,    and    a  =  0.19635  d\ 

Example.  A  cast-iron  ball  of  d  =  8  inches  in  diameter,  weighing 
W=  69.88  pounds  or  M  =  2.1  matts,  is  fired  from  a  gun  with  a  ve- 
locity V=  1000  feet  per  second.  Required  the  time  t  in  which  the 
ball  will  reach  a  target  at  8'  =  1500  feet  horizontal  distance  from  the 
muzzle  of  the  gun  ? 

a  =  0.19635  x  8J  =  12.5664  square  inches. 
12.5664  1 


Coefficient     C= 


122367        9737.65 


HypJcg.  2.37  =  0.815253,    and    X-  M-  0.27. 

9737.65x0.27     , 
Time     t  = =  2.629  seconds. 


168  ELEMENTS  OF  MECHANICS. 


In  this  time  the  ball  would  fall 

32.17x2.629' 


... 
=  1H.17  feet. 


Then,  in  order  to  hit  the  mark  on  the  target,  the  centre  line  of  the 
gun  must  be  pointed  to  111.17  feet  above  that  mark,  or  the  gun  must 
be  elevated  to  an  angle 


=tanA°  15'. 


1500 


§97.    RESISTANCE   OF  THE   AIR   WHEN   THE   BODY   DESCRIBES   A 
PARABOLA. 

When  the  gun  is  elevated  to  an  angle  z  and  the  ball  describes  a 
parabolic  curve  a  c  d,  the  horizontal  velocity  v  =  V  cos.z,  which  in- 
Fig.  127.  serted  for  Fin  the  preceding  formulas, 

will  make  them  answer  for  this  case 
also.  The  vertical  action  of  resist- 
ance in  the  ascent  is  counteracted  in 
the  descent,  so  that  only  the  horizon- 
tal resistance  of  the  air  need  be  con- 
sidered in  the  operation. 

Without  resistance  of  the  air  the 
ball  would  describe  the  parabola  a  ef, 
and  make  the  horizontal  range  S=Sr  +s,  but  the  resistance  diminishes 
that  space  by  s  or  to  S'  =  S-  s. 

MEAN    FORCE. 

A  variable  force  acting  on  a  body  free  to  move  can  be  converted 
into  a  mean  force. 

The  mean  force  in  time  is  the  mean  force  of  momentum. 

The  mean  force  in  space  is  the  mean  force  of  work. 

The  mean  force  in  time  is  at  the  centre  of  gravity  of  the  momentum 
area  of  the  base. 

The  mean  force  in  space  is  at  the  centre  of  gravity  of  the  work 
volume. 

I  98.     MEAN  FORCE  #  IN  TIME  T.     Figs.  121  and  122. 

When  a  force  varies  directly  or  inversely  as  the  time,  the  mean 
force  in  the  time  T  is  $  =  \F,  and  in  the  time  t,  $  =  $(F+f). 


MEAN  FORCE.  169 


The  mean  force  #  acting  on  a  body  in  the  time  Twill  produce  the 
same  velocity  as  that  of  the  variable  force  in  the  same  time. 

The  momentums  of  time,  #  T=  %  F  T=  M  V,  the  momentum  of 
motion. 

|  99.     MEAN   FORCE   <P   IN   SPACE   S.     Fig.  123. 

When  a  force  varies  directly  .as  the  time  of  action  the  mean  force 
in  the  space  S  is  #  =  f  F,  and  in  the  space  s  #  =  %(F+f). 

The  mean  force  acting  through  the  space  S,  will  store  the  same 
work  in  the  body  as  that  of  the  variable  force  in  jhe  same  space. 

I  100.     MEAN  FORCE  <i>  IN   SPACE   ,S.     Fig.  122. 

When  the  force  varies  inversely  as  the  time,  the  mean  force  #  in 
the  space  8  will  be  as  follows : 

Work,  $S=%MV\ 

Insert  Formula  5,  §  91,  for  8  and  Formula  3  for  V. 
^C  T3 


Work, 


SM 
<?T    3.F 


Mean  force,      #=f.F. 

That  is  to  say,  when  the  force  varies  either  directly  or  inversely  as 
the  time  of  action,  the  mean  force  in  the  space  passed  through  is 
*-4JE 

Work,       *S-II'8±IMV*. 

\  101.     MEAN    FORCE   0   IN   SPACE   *V.     Figs.  123  and  124. 

When  a  force  varies  directly  or  inversely  as  the  space  passed 
through  the  mean  force  in  the  space  8  is  $  =  %  F,  and  in  the  space 

ayf-JtfV). 

Work,        0  S  =  £  F  8-  \(F+f}s  =  \MV\ 

%  102.     MEAN    FORCE   <I>   IN   TIME   T.     Fig.  123. 

When  a  force  varies  directly  as  the  space,  the  mean  force  0  in  the 
time  T  is  found  as  follows : 

Momentum  of  time  #  T=  M  V,  momentum  of  motion. 

15 


170  ELEMENTS  OF  MECHANICS. 

Insert  Formula  1,  §  91,  for  T  and  Formula  2  for  V. 


Mean  force, 


hyp.log.S 


\  103.     MIAN    FORCE    <P   IN   TIME    T.     Fig.  124. 

When  the  force  varies  inversely  as  the  space,  the  mean  force  #  in 
the  time  T  is  found  as  follows  : 

Momentum  of  time  $  T=  M  V,  momentum  of  motion. 
Insert  Formula  3,  §  92,  for  T  and  Formula  2  for  V. 


Mean  force,  <P=CS. 


§104.    VIS-VIVA. 

Vis-viva,  literally  translated,  means  living  force  ;  the  term  is  used 
to  denote  double  the  work  stored  in  a  moving  body,  or  M  V1. 

Force  of  any  kind  is  an  element,  and  should  not  denote  work,  which 
is  a  function.  The  term  Vis-viva  conveys  an  idea  which  has  often 
been  entertained  —  namely,  that  a  dead  body  can  possess  a  virtue  of 
life  ;  which  erroneous  notion  has  caused  much  discordance  in  the  eluci- 
dation of  dynamics. 

Any  change  of  motion  in  a  body,  whether  from  rest  to  motion  or 
from  motion  to  rest,  requires  the  same  kind  of  force  —  namely,  that 
opposing  the  force  of  inertia,  which  should  not  be  termed  living  force. 

In  order  to  form  a  clear  conception  of  this  mysterious  Vis-viva,  we 
shall  bring  it  bodily  in  sight,  so  that  we  can  look  at  it. 

In  the  accompanying  illustration,  Fig.  128,  the  prism  a',  V  ,  </,  d",J 
is  the  same  as  that  in  Fig.  113,  which  represented  the  work  consumed 
in  bringing  the  moving  mass  M  of  velocity  V  to  rest.  In  the  base  of 
this  prism  we  have  two  elements,  force  F"  and  time  T'  ,  the  product 
of  which  makes  the  rectangle  a',b',f',(f  the  momentum  of  time.  The 
height  of  the  prism  represents  the  velocity  V  of  the  mass  M. 


VIS -VIVA. 


171 


Continue  the  line  a'  b'  to  /,  and  make  b'  f=b'  f .  Draw  through 
a'  the  diagonal  f  g  and  complete  the  parallelogram  f,f ,  g,  h;  continue 
the  line  b'  a'  to  b,  then  the  line  a'  b  re- 
presents the  mass  M  of  the  moving  body, 
and  the  line  a'  i  =  b  h  represents  the  ve- 
locity V.  The  parallelogram  a',  b,  h,  i 
is  then  the  momentum  of  motion  M  V, 
which  is  equal  to  the  parallelogram  a',  b',' 
f,  e,  the  momentum  of  time,  or  F  T 
-MV. 

Draw  the  line  h  d=  Fand  complete  the 
parallelopipedon  a',  b,  c,  d,  e,  i,  which  will 
then  represent  M  F2,  the  so-called  Vis- 
viva,  which  is  a  function  of  matter,  and 
motion. 

The  cubic  content  of  the  parallelopipe- 
don M  V1  is  double  that  of  the  prism 

which  represents  the  work  of  bringing  the  mass  M  of  velocity  V 
to  rest. 


Fig.  129. 


|  105.     VIS-VIVA   OR   LIVING   FORCE   IN    A   BODY   AT   REST. 

Fig.  129  represents  a  platform  car  attached  to  a  locomotive  run- 
ning on  a  railway  track  with  a  velocity  V.  On  the  fore  end  of  the 
platform  is  placed  a  body  or  mass  M  moving  with  the  car,  but  is 
stationary  in  relation  to  the  moving  system.  A  force  F  is  applied  on 
the  body  M  to  move  it  backward  until  it  strikes  the  obstruction  A 
with  a  velocity  V  equal  to  the 
foreward  motion  of  the  car.  The 
body  will  then  strike  the  ob- 
struction A  with  a  living  force 
or  Vis-viva  M  V2. 

Now  remove  the  obstruction 

A  and  place  the  body  M  in  its  original  position  at  a,  and  apply  the 
same  force  F  to  move  the  body  backward,  so  that  it  will  attain  the 
same  velocity  V  at  b ;  as  there  is  no  obstruction  at  A,  the  body  will 
fall  down  on  the  track,  where  it  will  lie  perfectly  still,  but  full  of  liv- 
ing force,  or  the  Vis-viva  stored  into  it  by  the  force  F  on  the  plat- 
form car. 

In  the  first  case  the  Vis-viva  of  the  body  was  discharged  by  the 
obstruction  A,  but  in  the  second  case  the  obstruction  was  removed, 
and  no  discharge  of  Vis-viva  has  taken  place. 

This  illustration  is  inL<  nded  to  prove  that  there  is  no  more  Vis-viva 


172 


ELEMENTS  OF  MECHANICS. 


Fi«-  13°- 


in  a  body  when  in  motion  than  when  at  rest.  The  body  M  was 
originally  set  in  motion  with  the  car  until  it  met  the  resistance  F, 
which  brought  the  body  to  rest  at  b,  but  the  obstruction  A  set  it  in 
motion  again  with  the  car,  and  in  the  last  case,  when  the  obstruction 
was  removed,  there  was  no  force  to  set  it  in  motion  with  the  car,  for 
which  reason  the  body  remained  at  rest  on  the  track. 

|106.     ON    RAISING   OR   LIFTING   A   BODY   VERTICALLY. 

The  illustration  represents  a  weight 
W  to  be  raised  from  a  to  d  by  a  force 
F  =  W,  or  the  force  will  just  balance 
the  weight.  The  force  is  represented 
to  push  the  weight  upward,  the  effect 
of  which  is  the  same  as  if  applied  in 
the  eye-bolt  to  lift  the  weight  ;  but  the 
illustration  is  clearer  by  representing 
the  force  to  push. 

The  force  F  only  balances  the 
weight  W,  and  can  thus  not  produce 
motion,  for  which  an  additional  force 
/  is  required.  The  additional  force/ 
is  applied  only  in  the  height  a,  b,  or 
space  s  in  the  time  t,  in  which  it  pro- 
duces the  velocity  V,  or 


d_ 


CL 


F*/ 


W 


The  acceleratrix  g  is  inserted  to  convert  the  weight  T'Finto  mass. 
After  the  force  /  has  ceased  to  act,  the  weight  W  will  continue  with 
*.he  force  F  in  a  uniform  ascending  velocity  until  some  other  force  is 
applied  to  retard  and  finally  stop  the  motion.  When  the  weight  has 
reached  the  height  c,  the  force  F  is  diminished  by  a  force  /',  which 
portion  of  the  force  of  gravity  of  the  weight  W  acts  to  retard  the 
motion  and  bring  the  weight  to  rest  at  d.  The  force  /'  acted  only  in 
the  space  s'  and  time  t  ',  so  that  the  momentums  of  time  ft  =/'  t,  or 
the  momentum  of  time  consumed  in  setting  the  weight  in  motion  from 
«  to  b,  is  re-utilized  from  c  to  d  .  There  is  no  relation  between  the 
forces  f  and  /',  which  are  entirely  dependent  upon  their  respective 
times  of  action. 

In  the  diagram  of  work  the  shaded  prism  /  Vt  represents  the 
primitive  work  consumed  in  setting  the  weight  in  motion  from  a  to  b, 
which  is  equal  to  the  shaded  prism/'  Vt!,  or  the  realized  work  of 
inertia  which  is  contributed  to  that  of  raising  the  weight  from  c  to  d. 


RAISING  A    WEIGHT.  173 

Therefore,  the  work  consumed  in  raising  a  weight  W  a  vertical  space 
S  is  equal  to  the  product  of  the  weight  and  space,  or 

Work     K=  W8. 

It  appears  in  this  formula  that  work  is  independent  of  time  and 
velocity,  because  the  work  will  be  the  same  for  whatever  time  occu- 
pied in  raising  the  weight,  and  also  with  whatever  velocity  it  is  raised  ; 
but  the  space  S  cannot  be  generated  without  its  constituent  elements 
—time  and  velocity  ;  and  either  one  of  these  two  elements  can  vary 
only  at  the  expense  of  the  other,  so  that  their  product  gives  the 
space,  or  S-  V  T. 

No  work  can  be  accomplished  without  either  one  of  its  constituent 
elements  F  V  T.  Either  one  or  two  of  the  elements  can  vary  ad  lib- 
itum, but  only  at  the  expense  of  the  remaining  two  or  one. 

The  forces  /and/  correspond  with  the  forces  jp'and  I*  ,  Fig.  113. 

Example.  In  reference  to  the  illustration  Fig.  130,  we  may  assume 
the  force  F  or  weight  TF=50  pounds,  and  the  additional  for  /=  5 
pounds  acting  in  a  time  of  t  =  1  second.  Kequired  the  velocity  V? 


F_*  =  i21Txl 

2  2 

The  body  is  now  ascending  with  the  uniform  velocity  V=  3.217  feet 
per  second  for  a  time,  say  T=  3  seconds,  which  will  make  a  space  of 
3x3.217  =  9.651  feet,  when  the  force  F  is  diminished  with/  =10 
pounds.  Required  the  time  if,  in  which  the  body  will  be  stopped? 

WV    50x3.217 

Time  '== 


The  space     s'  -       -  =  .  0.804  of  a  foot. 

The  whole  operation  of  lifting  the  weight  was  accomplished  in 
Time.  Space. 

t  =  1     second.  s  =  1.6085  feet. 

T=3     seconds.  6c  =  3.651    feet. 

f  -  0.5  seconds.  s'  =  0.804    feet. 

t  +  T+  1  =  4.5  seconds.  £=  6.0635  feet. 

The  work  in  lifting  the  weight  will  be 

K=  50  x  6.0635  =  303.175  foot-pounds. 

15* 


174 


ELEMENTS  OF  MECHANICS. 


§  107.     ON    IRREGULAR    WORK    IN    WHICH    BOTH    FORCE  AND  VE- 
LOCITY  ARE   VARIABLE. 


Fig.  131. 


Assume  the  case  of  a  body  M  moving  with  a  velocity  V  against 

a  spring,  which  force  of 
elasticity  is  ^Pwhen  com-* 
pressed  the  space  8. 

It  has  before  been  stated 
that  the  elastic  force  of 
spiral  springs  is  directly 
as  the  compression. 

The  mean  force  in  the 
space  S  will  then  be  %  F, 
or  the  mean  force  at  any 

compression  s  will  be  — — -. 

p1  Of 
The  work  done  in  the  compression  S  will  be  K= .  .     1 

2 

The  work  done  in  the  compression  s  will  be  k  =  — - .  .2 


The  work  consumed  on  the  moving  mass  is 


MV* 


When  the  mass  has  compressed  the  spring  the  space  s,  its  velocity 
will  be  reduced  to  v,  and  the  remaining  work  required  to  bring  the 

MV* 
mass  to  rest  will  be  -         .  .     4 


Therefore,  the  work  done  in  compressing  the  spring   the  space 

will  be  i^  =  KI!_M^  5 

'  ' 


— 


~ 


This  formula  gives  the  velocity  v  with  which  the  mass  M  com- 
presses the  spring  at  the  space  s. 


WORK  IN  A  SPRING.  175 

The  figure  under  the  spring  represents  the  nature  of  the  work  per- 
formed in  compressing  the  spring,  supposing  that 

2K-M  V-F8.      ....     7 

The  vertical  lines  represent  the  velocity,  and  the  dotted  lines  in 
the  base  the  force  of  the  spring  at  the  corresponding  compression  s. 

The  power  in  operation  at  any  compression  s  is  represented  by  the 
section  y  v  of  the  figure. 

The  cubic  contents  of  the  figure  represents  the  work. 

Example  1.  It  is  found  by  experiment  that  the  spring  can  be  com- 
pressed S  =  1  foot  by  a  force  F=  360  pounds.  Required  the  velocity 
of  the  mass  M  =  12  matts.  to  compress  the  spring  one  foot  ? 


M  V*  =  FS,         V=\=  _  5.477  feet  per  8econd> 

\  Jjf      \      12 

Example  2.      Required  the  velocity  t;  at  the  compression  s  =  0.5 
of  the  foot? 


Formula  6.     v  =  x/5.477'  -  360xQ-5*  =  4.  75  feet  per  second. 
»  12  x  1 

The  force  of  the  spring  at  this  velocity  and  compression  is 

,    Fs     360x0.5     . 
/=  --  =  -  -  =  180  pounds, 


and  the  power  in  operation  at  that  moment  is 

855 

p  =/  v  =  4.75  x  180  =  855  effects,     or    -  =  1.55  horse-power. 

550 

Example  3.  A  mass  M  =  6  matts.  moving  against  the  spring  with 
a  velocity  F"=4  feet  per  second.  Required  its  velocity  when  the 
spring  is  compressed  s  =0.6  feet? 


Formula  6.     v  =  -y/4* ; — -1 —  =  -  2.37  feet  per  second. 


The  negative  sign  proves  that  the  mass  was  not  able  to  compress 
the  spring  s  =  0.6,  for  which  is  wanted  an  additional  velocity  of  2.37 
feet  per  second  to  V]  or  F"=6.37  feet  per  second  would  just  compress 
the  spring  that  much  but  no  more,  and  the  velocity  v  would  then  be  0. 


176 


ELEMENTS  OF  MECHANICS. 


\  108.    THE  MASS  ACTED  UPON  BY  TWO  OPPOSING  FORCES. 

The  force  .Facts  on  a  mass  M sliding  on  the  base  B  with  friction/. 
The  mass  is  thus  acted  upon  by  two  opposite  forces,  Fand/  of  which 

Fig.  132. 


Fraust  be  greater  than /before  the  mass  can  be  moved,  and  the  force 
which  sets  the  mass  in  motion  is  F—f.  The  force  F  acts  on  the  mass 
only  in  the  space  S,  when  the  work  of  the  force  will  be 

K-FS. 1 

The  work  done  by  the  friction  will  be 

k~fS. 2 

The  work  utilized  in  giving  the  mass  M  the  velocity  V  in  the  space 
8  will  be 


of  which 


M:(F-f)=T:V.     . 

MV    2S_ 
F-f     V  ' 


M 


"V       M       '      '        '        ' 


FRICTION  FORCE.  177 


of  which  T=. 8 

s_T(F-f)_    MV* 
2  M       2(F-f) 


.  10 
MV    , 


.11 


The  prism  F  V  T  represents  the  total  work  K  of  the  force  F  and 
Formula  1,  of  which  the  work  consumed  by  the  friction  is  repre- 
sented by  the  light  part  of  the  prism  fV  Tand  Formula  2. 

The  shaded  part  of  the  prism  represents  the  work  utilized  in  giving 
the  mass  M  the  velocity  V,  which  corresponds  with  Formula  3. 

The  second  prism  fVt  represents  the  work  of  friction  consumed  in 
bringing  the  mass  to  rest  after  the  force  F  ceased  to  act.  The  cubic 
content  of  the  prism  fVt.  is  equal  to  that  of  the  shaded  part  of  the 
first  prism  which  set  the  body  in  motion. 

Example  1.  The  force  F=  160  pounds,  Jf=64  matte.,  and  the 
friction  /=  100  pounds.  What  time  is  required  to  move  the  mass  M 
a  space  $=20  feet?  and  what  will  be  the  velocity  F?  at  the  end  of 
that  space  ? 


_,         .    _      T,        2x20(160-100)     C10,  ,    ,  , 

Formula  7.      V=  */—  —  =  6.125  feet  per  second, 


the  required  velocity  at  the  end  of  the  space  S. 
Formula8.     r=™_4 

the  time  required  to  move  the  body  S=2Q  feet. 
M 


178  ELEMENTS  OF  MECHANICS. 


I  109.     ON  FRICTION  OF  SCREW  PROPELLERS  WORKING  IN  WATER. 

The  friction  of  screw  propellers  running  in  water  is  a  considerable 
item  of  the  propelling  power,  and  is  well  worthy  of  attention  when 
speed  of  the  vessel  and  economy  of  fuel  are  desired.  The  solution  of 
the  problem  furnishes  a  good  example  of  the  value  of  the  calculus  in 
dynamics. 

Fig.  133. 


Notation  of  Letters. 

P=  pitch  of  the  propeller  in  feet. 
R  =  radius  of  the  propeller  in  feet. 

r  =  any  radius  less  than  R. 

1  =  length  of  a  helix  for  one  whole  convolution  in  feet,  at  the 

radius  r. 
A  =  area  of  the  helicoidal  surface  for  one  whole  convolution  in 

square  feet. 
jV=  number  of  blades  of  the  propeller. 

n  =  number  of  revolutions  per  minute  of  the  propeller. 

v  =  velocity  of  the  helix  I  in  feet  per  second. 

h  =  horse-power  of  friction. 

f=*  friction  in  pounds. 

£  =  differential. 

The  friction  in  pounds  per  square  foot  of  surface  of  cast-iron  and 
brass,  of  rough  castings,  and  also  of  smooth  surfaces,  filed  or  ground, 


FRICTION  OF  SCREW-PROPELLERS.  179 

but  not  polished,  is  approximated  as  follows  when  the  surface  moves 
with  the  veldrtty  of  one  foot  per  second  : 

Friction  Surface.  /' 

Rough  cast-iron 0.0045 

Smooth  cast-iron : 0.0040 

Brass,  rough  casting ;JT^5.0040 

Brass,  smooth' surface 0.0030 

The  friction  for  rough  cast-iron  will  then  be,  in  pounds, 

/=  0.0045  Av* 1 

The  differential  area  of  the  helicoidal  surface  for  one  convolution 
will  be 

8  A  =  1 cr,    .        .        .         .        .2 

and  the  differential  friction  in  pounds  will  be 

»/=  0.0045  v*ldr 3 

This  force  of  friction  multiplied  by  its  velocity  will  be  power  in 
effects,  and  divided  by  550  will  be  horse-power,  when  the  differential 
horse-power  will  be 

a,     0.0045  v3lcr 

en  = .       .         .  .4 

550 

In  ,     Pn* 

The  velocity  v  =  —    and    ^  =*-— .         -•        .        .     5 

The  differential  horse-power  will  then  be 
0.0045  n^Er 
550x60* 

-r    0.0045  wl  n3 

Call  X= = .  7 

550  xGO3      26,400,000,000 


Then                                   £A  =  XJ40r,          ....     8 
and  k  =  xf?dr 9 

But  J-1/4«»rl  +  Pa>     and     P  =  (4  **  r2  +  P2)2.  .  10 

Insert  Formula  10  in  Formula  9,  and  we  have 

.  *     .         .11 
4.      .         .12 
Then  h  =  X T(16  ^  r*  +  8  ^  r*  P2  +  P4)  Cr.        .         .13 


180  ELEMENTS  OF  MECHANICS. 

By  integrating  each  term  in  the  parenthesis  we  have 
16  *4  r5    8  7T2  7*  * 


and  =  26.319  r>  P'.      .15 


Integrate  the  friction  horse-power  from  the  centre  of  the  propeller 
to  the  periphery  of  radius  P.,  then  when  r  =  0,  C=  0. 

Insert  the  Formula  7  for  JTin  Formula  14,  with  the  values  15,  and 
we  have  the  frictional  horse-power  for  one  convolution  of  the  screw 
and  for  one  side  of  the  helicoidal  surface, 


A-L{/ /qi  1  71    7?*_u9AQ1Q   7?*   £>z-i-  P4N\  Ifi 

=  — |  oil.  1 1  .fir  +  ^O.Oiy  s±    J:    -T  J:    ).         .ID 

26,400,000,000 

The  helicoidal  Surface  of  screw  propellers  is  generally  cut  up  into 
small  portions  by  a  number  of  blades,  each  of  a  fraction  of  the  pitch, 
and  when  the  helicoidal  surface  is  counted  on  both  sides  of  the  blade, 
the  friction  horse-power  will  be 


-  _  —  P/+26.319  P'P'+P4)      .17 

13,200,000,000  P^ 

This  formula  includes  both  the  dragging  and  rotary  friction  horse- 
power of  a  propeller  of  rough  cast-iron. 

Call/'  =  friction  in  pounds  per  square  foot  of  surface  moving  with 
a  velocity  of  one  foot  per  second,  and  the  friction  horse-power  will  be 


Example.  Required  the  friction  horse-power  of  a  propeller  of  the 
following  dimensions: 

Diameter  of  propeller,  20  feet,  or  ............  P  =    6  feet. 

Pitch  ................................................  P=  18  feet. 

Length  in  the  direction  of  axis  ...............  Z  =  2.4  feet. 

Number  of  blades  ................................  N=   4. 

Revolutions  per  minute  .........................  n  =  60. 

The  horse-power  consumed  by  friction  will  then  be,  for  rough  cast- 
iron  surface, 

. 


59,400,000x18 
horse-power  of  friction. 


FRICTION  OF  SCREW-PROPELLERS.  181 

In  the  year  1850  an  experimental  steamer  was  built  in  Kensington, 
Philadelphia,  which  was  expected  to  make  20  to  30  miles  per  hour. 
The  propeller  was  about  4  feet  in  diameter,  with  only  one  blade,  extend- 
ing the  whole  convolution  of  the  circle,  and  with  a  very  fine  pitch  of 
about  6  inches.  (The  author  does  not  remember  the  exact  dimensions.) 
The  propeller  was  expected  to  make  500  revolutions  per  minute. 

Example.  —  Required  the  friction  horse-power  of  the  above-de- 
scribed propeller.  12  =  2  feet,  P  =  0.5  feet,  n  =  500,  L  =  0.5,  N=l. 
Surface,  rough  cast-iron. 

500-  2  2  6 


59,400,000x0.5 
horse-power,  nearly. 

The  power  of  the  engine  counted  from  the  size  of  the  steam  boiler, 
did  not  amount  to  more  than  50  horse-power,  and  the  result  was  that 
when  the  trial  trip  came  off  the  steamer  could  hardly  crawl  up  against 
the  tide. 

The  building  of  the  steamer  was  kept  in  the  greatest  secresy,  and 
her  performance  was  expected  to  astonish  the  world. 

There  was  another  experimental  steamer  built  in  Kensington  in  the 
year  1864,  in  which  several  curious  propellers  were  placed  on  each 
side  of  the  vessel,  which  also  turned  out  a  failure  on  account  of  the 
friction  of  the  propellers  in  the  water  being  too  great. 

A  fine-pitched  propeller  has  more  friction-power  than  one  with 
sharp  pitch  for  equal  speed  of  vessel. 

The  proper  pitch  for  propelling  steamboats  should  be  between  two 
and  two  and  a  half  times  the  diameter  of  the  propeller.  The  sharpest 
vessel  should  have  the  sharpest  pitch  of  propeller. 

For  the  proper  proportions  and  construction  of  screw-propellers, 
see  "  Nystrom's  Pocket-Book  of  Mechanics,"  thirteenth  edition. 
16 


182  ELEMENTS  OF  MECHANICS. 

§110.    GYRATION. 
Gyration  means  Circular  Motion. 

The  term  is  used  in  dynamics  of  matter  in  circular  motion  to 
designate  the  mean  effect  of  all  the  particles  in  a  revolving  body  or 
system  of  bodies. 

In  motion  of  translation  all  the  particles  of  a  body  move  with 
equal  velocity  in  straight  and  parallel  lines,  but  in  circular  motion 
tie  particles  move  with  different  velocities  in  different  circles,  of 
which  one  is  called  the  circle  of  gyration. 

$  111.     DYNAMICS  OF   MATTER   IN   CIRCULAR  MOTION. 

Fig.  134  represents  a  mass  M  free  to  move  around  the  centre  O  in 
the  circle  of  radius  H.     A  constant  force  F  is  applied  in  the  direc- 
Fig.  134  **on  °^  ^e  tangent  of  the  circle  to  move  the 

mass,  the  dynamics  of  which  will  be  the  same 
as  if  the  mass  moved  in  a  straight  line  or 
•«  -^    ^  I?  rr> 

***  /£> 


The  velocity  V  of  the  mass  M  will  be  in 
the  direction  of  the  circle,  which  periphery 
is  2  *  R. 

Circular  velocity  is  generally  expressed  in  revolutions  per  minute, 
and  denoted  by  the  letter  n.  When  the  radius  JR  is  expressed  in  feet 
the  circular  velocity  -mil  be 

„    2  -  R  n  60  F 


Example  1.  A  force  F=  36  pounds  is  applied  on  the  mass  J/=  48 
matts.  for  a  time  T=  9  seconds.  Required  the  velocity  F?  and  revo- 
lutions per  minute  ?  with  which  the  mass  will  continue  to  rotate  after 
the  force  ceased  to  act  ?  The  radius  of  the  circle  being  R  =  2  feet. 

of>      q 

=  6.75  feet  per  second. 


48 


- — - — : =  32.6  revolutions  per  minute. 

i-  X  t>.14  X  ^ 


GYRATION..  183 


The  mass  M  will  continue  to  rotate  with  this  velocity  until  some 
force  F'  is  applied  in  opposite  direction  to  retard  the  motion  and 
finally  bring  the  mass  to  rest. 

Example  2.  What  force  F'  is  required  to  stop  the  rotation  of  the 
mass  Miu  a  time  of  jT=4  seconds? 

_,     MV    48x6.75 
F'  =  —  = =  80  pounds. 

The  primitive  work  consumed  in  setting  the  body  in  rotation  is 
equal  to  the  realized  work  by  which  the  rotation  is  stopped,  like  in 
straight  linear  motion,  the 


2  2  x  602 

=  0.00548314  MJPri*,  the  work  of  rotation. 

Io.2.o77 

This  formula  gives  the  primitive  or  realized  work  of  a  rotating 
mass,  as  illustrated  by  Fig.  134. 

When  the  mass  is  expressed  by  weight  W,  the  formula  will  be 

=  32.17x182.377  =  5867.16  ' 

When  the  force  F  is  applied  on  a  radius  r,  and  the  centre  of  the 
mass  M  rotates  with  a  radius  £,  as  represented  by  Fig.  136,  the  for- 
mulas will  be  as  follows  : 

F=^T"        "       <* 


M_Fr  T 

~  EV' 

v=Fr  T 
=  ME' 

T= 


2 


3 


FT    ' 


^   *n  ^    ,,, 

60  rT    ' 


no  Fr  T 


GO  Fr  T 


n  = 


6 


7 


^  .< ..  ^  „ 
60  Fr 


Example  1.  A  mass  J[f=96  matts.  is  to  be  put  into  a  circular 
motion  of  n  =  360  revolutions  per  minute  around  a  radius  R  =  3  feet. 
The  force  F  which  sets  the  mass  in  rotation  acts  on  a  lever  or  crank 
of  radius  r  =  0.5  of  a  foot.  What  force  is  required  to  give  the  mass 
that  circular  velocity  in  a  time  T=*  10  seconds  ? 

,    96x2x3.14x3'x360 

Formula?     jF= =  6511.1  pounds. 

60x0.5x10 


184  ELEMENTS  OF  MECHANICS. 


§112.     CENTRE   OF   GYRATION. 

Centre  of  gyration  in  a  revolving  body  is  a  point  in  which,  if  all 
the  matter  were  there  contained,  it  would  have  the  same  dynamic 
effect  as  when  distributed  around  that  centre. 

The  centre  of  gyration  describes  the  circle  of  gyration. 

The  centre  of  gyration  is  always  outside  of  the  centre  of  gravity 
of  the  revolving  body. 

The  less  space  the  body  occupies  in  the  circle  of  gyration,  and  the 
greater  the  radius  of  revolution  is,  the  nearer  does  the  centre  of  gy- 
ration approach  the  centre  of  gravity  of  the  body. 

|113.    RADIUS   OF   GYRATION. 

The  radius  of  gyration  is  the  distance  from  the  centre  of  rotation 
to  the  centre  of  gyration  in  a  revolving  body. 

The  radius  of  gyration  will  herein  be  denoted  by  the  letter  X,  to 
distinguish  it  from  other  radii. 

§114.     MOMENT   OF   INERTIA. 

Moment  of  inertia  is  either  weight,  mass,  volume  or  surface,  multi- 
plied by  the  square  of  its  radius  of  gyration. 

The  moment  of  inertia  will  hereafter  be  denoted  by  the  letter  E, 
and  may  be  expressed  in  either  of  the  following  units : 

E  =  W  X*  in  square  foot-pounds,  or  sq.  ft.  Ibs. 
E=  MX*  in  square  foot-matts.,  or  sq.  ft.  mts. 
E=  Q  X*  in  square  foot-volumes,  or  sq.  ft.  vol. 
E=  0  X'  in  square  foot-square,  or  sq.  ft.  sq. 
Moment  of  inertia  is  used  for  finding  the  radius  of  gyration. 

P5    135  Let  Q  denote  the  volume  of  any  system  of  bodies, 

and  X  its  radius  of  gyration  rotating  around  the 
centre  o,  and  A,  £,  C,  D,  etc.  represent  the  several 
volumes  of  respective  radii  of  gyration  v,  x,  y,  z, 


i          Then     Q  X**=A  vt+£xt  +  Cy^+JJz*,  etc. 
The  radius  of  gyration  of  the  system  will  then  be 


,  etc. 


INERTIA.  185 


We  have  learned  that  the  work  of  a  revolving  body  is 


182.377 '  5867.16 ' 

in  which  R  means  the  radius  of  gyration,  MR'1  and  W R*  means  the 
moment  of  inertia. 

When  the  moment  of  inertia  E  is  given  the  work  in  the  revolving 
body  will  be 

^=  -lonor^'       OI>       ^= 


182.377'  5867.16' 

The  moment  of  inertia  is  a  constant  quantity  in  a  body  or  system 
of  bodies  rigid  to  an  axis  of  rotation. 

Moment  of  inertia  is  E=MX'*  .....     1 

Square  radius  of  gyration  is  X*  =  —  .    .         .         .         .         .2 

E 


Radius  of  gyration  is  then      X=  •*  /  —  . 

\  E 


E 

Mass  of  the  body  will  be        M=  —  -t.  . 

A. 


\  116.     FORCE    OF   INERTIA. 

The  force  of  inertia  is  equal  to  any  force  applied  to  change  the 
motion  of  a  body. 

Let  a  constant  force  jPbe  applied  on  a  lever  r 
to  change  the  rotary  motion  of  the  mass  M  re- 
acting with  its  force  of  inertia  ^Ton  the  lever  or 
radius  of  gyration  X. 

Then,  F-.I  =  X:r, 

and  the  static  momentums  F  r  =  I X. 

F  r 

The  force  of  inertia  will  be  I— . 

X 

Let  I7  denote  the  time  in  seconds,  in  which  the  velocity  of  the  mass 
J/  is  changed  from  V  to  v  feet  per  second,  or  from  N  io  n  revolutions 
per  minute. 

Then,  I:  M=(V-v):  T, 

lo  * 


186  ELEMENTS  OF  MECHANICS. 

from  which  the  force  of  inertia  will  be 

M(V-v)  =  Fr 
T      '    X  ' 

Static  momentum     Fr  =  IX=  MX(V~V\ 


^r  ,      .__      .  .  A_       N 

F=  --  -,    v  -  --  ,    and     (  V-  v)  =  -  (N-  »). 

60  60  60  v 

9  —  JUT  Y"s 

Static  momentum    '  F  r  =  /  X=  —  -  (TV-  w). 

The  time  T  in  which  the  angular  velocity  is  changed  from  V  to  v, 
or  revolutions  from  ^Tto  n,  will  be 


Fr  60  Fr 

The  change  of  angular  velocity  in  the  time  Twill  be 

Fr  T  ...      .       &OFrT 

(F-»)-—  ,    or    (a--,)- 

The  force  jP  required  on  the  crank  will  be 


MX(V-v) 


Tr  60  Tr 

Example.  A  body  weighing  2255  pounds,  or  mass  M=  70  matts., 
revolving  with  a  radius  of  gyration  X=  5  feet,  and  making  N=  120 
revolutions  per  minute,  is  acted  upon  by  a  force  F=  96  pounds  on  a 
lever  r  =  0.75  of  a  foot.  What  time  Tis  required  to  reduce  the  revo- 
lution to  n  =  60  per  minute  ? 


or  2  minutes  32.6  seconds,  the  answer. 

Example.  A  mass  M=  360  matts.  is  revolving  with  a  velocity  of 
F=  30  feet  per  second  in  a  circle  of  X=  12  feet  radius  of  gyration, 
when  a  force  ^=90  pounds  acting  on  a  crank  r  =  1.25  feet  is  applied 


RADIUS  OF  GYRATION.  187 

in  T=  180  seconds  to  increase  the  angular  velocity.     Required  the 
increased  velocity  F? 

90x1.25x180    , 
(F-')-      S60xl2      -"• 

F-v  =  4.1,     or      F=  4.1  +  30  =  34.1  feet  per  second, 

the  velocity  required. 

The  force  of  inertia  of  any  revolving  body  in  the  circle  of  gyration 
will  be 


Example.  A  fly-wheel  weighing  5400  pounds,  or  M=  167.84  matts., 
is  making  N=  120  revolutions  per  minute,  with,  a  radius  of  gyration 
JT=4.5  fe^t.  The  angular  velocity  of  the  wheel  is  to  be  reduced 
to  n  =  15  revolutions  per  minute  in  a  time  T=  90  seconds.  What 
force  I  must  be  applied  in  the  tangent  of  the  circle  of  gyration  to  re- 
duce the  revolutions  from  120  to  15  in  the  time  90  seconds  ? 

.    167.84  x2x  3.14  x  4.5.  ir 

-  60^90  -  (       "     ^  "      P°Un 

This  is  the  force   of  inertia  under   the   conditions  given  in  the 

example. 

§  116.     RADIUS  OF  GYRATION  AND  MOMENT  OF  INERTIA. 

Let  two  masses  M  and  m  rotate  around  a  common  centre  C  in  cir- 
cles of  gyration  of  radii  E  and  r,  and  the  two  masses  being  rigid  so 
as  to  rotate  with  a  common  angular  velocity.  Fig  ia? 

Then  the  moment  of  inertia  of  each  mass  will 
be  M  R*  and  in  r2,  and  of  the  two  masses 


E  =  (M+  m)JP 

From  this  formula  we  have  the  radius  of 
gyration  to  be 


x= 


J/+7W 


Let  Z  denote  the  distance  of  the  centre  of  gravity  of  the  two  masses 
from  the  centre  C,  and  we  have  the  static  momentum 


Z(M+m)  =  HR+mr,  of  which  Z~         ±?LI. 

M+m 


188  ELEMENTS  OF  MECHANICS. 

It  is  to  be  proved  that  the  centre  of  gyration  is  outside  of  the  cen- 
tre of  gravity  of  the  two  masses,  or  that  X>  Z. 

^     Mtf+mr*  z,_IMR+mr\* 

M+m  \      M+m    j 

MlP+mr1       (MR+mr? 

M+  m         (M+  m)(M+  m) 
Reject  the  common  denominator  M+m. 


Multiply  both  members  by  M+m. 

(M+m)(MR"+m  rt)>(MR+m  r)1. 
M*  R*+Mm  R*  +  Mm  S  +  m*  r*>M*  R*  +  2MR  m  r  +  m*  r\ 

Mm  R*  +  Mm  r>>2  MR  m  r. 
R  2  +  r*  >  2  R  r  ;  which  is  true,  because  if  R  >  r, 
the  two  squares  R^+r1  must  be  greater  than  two  rectangles  R  r. 

HI"-     GYRATION    OF    IRREGULAR    BODIES. 

To  find  the  radius  of  gyration  and  moment  of  inertia  of  an  irregular 
body  revolving  around  the  centre  C,  Fig.  138. 

Draw  concentric  circles  at  equal  distances  r  apart  to  represent 
cylindrical  sections  of  the  body  concentric  with  the  axis  C. 

The  first  and  last  division  should  be  only  ^  r. 
The  more  divisions  made  the  more  correctly  will 
the  radius  of  gyration  and  moment  of  inertia  be 
ascertained. 

The  area  of  each  cylindrical  section  a,  b,  c 
and  d  will  then  represent  the  mass  of  rotation  at  the  respective  section. 
Let  0  denote  the  sum  of  all  the  areas  of  the  sections,  and  X=  radius 
of  gyration  of  the  body.     The  sum  of  the  moments  of  inertia  of  the 
sections  will  be  equal  to  0  X1. 

0  =  a+b+c+d. 
Moments  of  inertia,  0  X*  =  a  r2+6(2  r)J  +  c(3  r)2+d(4  r)1. 


Rad.  gyration, 


EQUILIBRIUM  OF  GYRATION.  189 


|118.     EQUILIBRIUM   AND   DISTURBANCE   OF   GYRATION. 

A  body  or  system  of  bodies  revolving  around  an  axis  c  upon  which 
it  may  be  perfectly  balanced  in  regard  to  gravity,  but  when  rotating 
with  a  variable  velocity,  the  equilibrium  is  disturbed  in  the  axis  of 
rotation. 

Two  masses  M  and  m  revolving  around  their  common  centre  of 
gravity  o,  will  be  in  gyratic  equilibrium  only  Fi  tgg 

when   their   angular  velocity  is   uniform,  in  \ 

which  case  the  equilibrium  is  not  disturbed  in 
the  axis  of  rotation ;  but  when  the  angular 
velocity  is  irregular,  there  is  a  tendency  to 
change  the  axis  of  rotation,  and  the  equilibrium  ("^VT  *"  /  ' 

of  gyration  is  disturbed.  / 

Let  R  and  r  denote  the  respective  radii  of 

gyration  of  the  masses  M  and  m,  and  X=  radius  of  gyration  of  the 
system. 

Suppose  the  bodies  to  be  balanced  so  that  M:m=r:R,  or 
M R  =  mr,  in  which  case  their  dynamic  momentum  will  be  alike,  or 
M  V=  m  v  ;  but  the  work  in  each  body  will  not  be  alike. 

From  §  112  we  have  the  works  in  the  revolving  bodies  to  be 

M&rf  mr^n* 

182.377  '    '  "  182.377  ' 

Insert  MR  in  the  work  k,  and  m  r  in  the  work  K,  which  will  be 

,,     m  r  R  n 

K= .    and    £  = 


182.377  '  182.377 

182.377  K     182.377  k 


mr  R  MRr 

KMRr^kmrR,    or    K:lc  =  m:M. 

That  is  to  say,  the  work  in  the  bodies  are  inversely  as  the  masses ; 
therefore,  the  work  which  sets  the  unequal  masses  into  rotation  is 
unequally  divided  in  the  centre  of  rotation,  which  causes  disturbance. 

When  the  masses  rotate  with  a  uniform  velocity,  there  is  no  work 
transmitted  to  or  from  them  through  the  centre  of  rotation,  and  there 
can  consequently  be  no  disturbance. 

Any  change  in  the  angular  velocity  will  cause  a  disturbance  in  the 
journals,  which  disturbance  cannot  be  avoided  except  by  making 
M-R-^mr2;  but,  then  the  rotating  system  will  cause  a  disturbance 
during  uniform  velocity,  because  it  is  then  not  balanced  for  gravity. 


190  ELEMENTS  OF  MECHANICS. 


§119.     GYRATION   TREATED    BY   THE   CALCULUS. 

Any  body  or  system  of  bodies  may  be  considered  to  be  composed 
of  an  infinite  number  of  cylindrical  sections  concentric  with  the 
common  axis  C  of  rotation. 

Let  0  denote  the  area  of  any  such  section 
rotating  around  the  axis  o,  with  the  radius  r. 
Q  =  cubic  content  of  the  body. 
Then  cQ  =  08r. 

Moment  inertia   Q  X*  = f  Or'dr+C. 
Radius  gyration      ^=\l  I +  ^' 

In  order  to  bring  the  formulas  into  a  practical  shape,  it  is  necessary 
to  know  the  variation  of  the  section  0  in  relation  to  r. 


^120.     GYRATION    OF   A   STRAIGHT   LINE   OR    ROD. 
Centre  of  Gyration  and  Moment  of  Inertia. 

Let  a  straight  line  or  thin  parallel  rod  L  be  attached  with  one 
p.    H1  end,  and  at  right  angle  to  the  axis  of  rotation  o  o. 

As  the  rod  is  parallel,  its  section  0  at  any  dis- 
tance I  is  a  constant  quantity. 


-1---X *. 

Moment  inertia    Q  X'1  =J  0^  U  =  %0  L3  sq.  ft.  vol. 

Q=OL,  and   0  =  — . 
L 

Moment  inertia   Q  X*  =  \Q  L*. 

IQ    Tt  IJTi 

Radius  gyration        X=  -^1 — —  =  ^1  —  =  L\/\. 

Radius  gyration       X=  0.5775  L. 

Example.  A  line  or  parallel  rod  is  L  =  12  feet  long,  and  revolves 
around  one  of  its  ends.     Required  the  radius  of  gyration  ? 

X=  0.5775x12  =  6.93  feet. 


GYRATION.  191 


§  121.  GYRATION  OF  A  BALANCED  LINE  OR  PARALLEL  ROD. 

The  radius  of  gyration  of  a  line  or  parallel  rod  revolving  around 
an  axis  passing  through  the  centre  of  gravity  Fig  142 

at  right  angle  to  the  line  will  be  o 

X=£x 0.5775,    L  =  0.28875  L.  ~  L\~  ' 

§122.     GYRATION   OF   A   STRAIGHT   LINE   NOT    EXTENDING   TO 
THE   AXIS   OF   ROTATION. 

When  the  revolving  line  or  parallel  rod  does 
not  extend  to  the  axis  of  rotation,  the  radius  of   0          mg' 
gyration  is  found  as  follows : 

The  radius   of  gyration  of  each   elementary 
section  of  the  line  or  rod,  will  be  (a  +  ljfjl. 

Moment  inertia  L  X*  =  f  (a+W  fil. 


=  f  (a+£)2 


S*a+l 

LX*=\  (a2 +2  a  l+l*)  fil  =  cf  L+a  Z?+%L3  =  L(o?-\ 
Radius  gyration    X=  yd1  +  a  L  +  \L'*. 

§  123.     CENTRE  OF  GYRATION  REFERRED  TO  CENTRE  OF  GRAVITY. 

Having  given  the  radius  of  gyration  of  a  body  or  system  of  bodies 
when  the  axis  of  rotation  passes  through  the  centre  of  gravity,  to 
find  the  radius  of  gyration  of  the  same  when  the  centre  of  gravity 
rotates  with  a  radius  R  around  an  axis  parallel  with  the  former  axis. 

Referring  to  the  preceding  figure,  the  radius  R  of  the  centre  of 
gravity  of  the  rod  will  be  R  =  a  +  %L. 

a  =  R-\L,    and    a'2  =  1? -  R  L  +  ^L*. 
The  radius  of  gyration  of  the  preceding  figure  is 


Insert  the  above  values  of  a  and  a2,  and  we  have 

x=  /'R 


Radius  gyration     X 

This  formula  gives  the  centre  of  gyration  for  any  parallel  figure  of 
length  L. 

""=  0.28875  L, 


192  ELEMENTS  OF  MECHANICS. 

which  is  the  radius  of  gyration  of  a  line  L  rotating  around  its  centre 
of  gravity.  Therefore,  if  r  =  radius  of  gyration  of  a  body  or  system 
of  bodies  rotating  around  its  centre  of  gravity,  and  R  =  radius  of 
centre  of  gravity  of  the  same  system  rotating  around  another  axis 
parallel  to  the  former,  then 

Radius  gyration     Jf=  j/.fif'-t-r2. 
This  formula  will  hold  good  for  any  form  of  body  or  system  of  bodies. 

g  124.     GYRATION  OF  A  BODY  REVOLVING   OUTSIDE    OF   ITS  AXIS 
OF   ROTATION. 

The  illustration  represents  the  plan  of  the  circles  of  rotation.     The 
body  A  B  is  first   supposed  to   revolve   around  its 
.  centre  of  gravity  o  when  its  centre  of  gyration  a  de- 

scribes the  dotted  small  circle  of  radius  r. 

Now  let  the  body  revolve  around  the  centre  o',  with 
a  radius  R  from  the  centre  of  gravity  o,  so  that  the 
radii  R  anc?  r  are  at  right  angles  to  one  another. 
Then  the  radius  of  gyration  will  be  X=  o'  a,  which 
is  the  hypotenuse  of  the  catheters  R  and  r,  and  the 
centre  of  gyration  will  in  this  case  be  at  b,  describing 
the  dotted  large  circle. 

This  is  an  illustration  of  the  formula  in  the  preceding  paragraph, 


The  body  A  JB  may  revolve  in  any  position  in  regard  to  the  circles 
of  gyration,  with  the  condition  that  the  two  axes  o  and  o'  must  be 
parallel. 

GYRATION  OF  A  RECTANGULAR  PLANE. 

The  radius   of  gyration  of   any  number   of 
parallel   lines   rigid   into   a  rectangular   plane, 
which  ends  are  parallel  with  the  axis  of    ro- 
tation, is  the  same  as  that  for  a  single  line. 
R  =  radius  of  centre  of  gravity  of  the  plane. 
r  =  radius  of  gyration  of  the  plane  when  the  axis  of  rotation 
passes  through  the  centre  of  gravity  and  is  parallel  to  the 
axis  o,  o. 
0  =  area  of  the  rectangular  plane. 

Moment  inertia     E~  0(Ri  +  r2)  =  0(R* 


Radius  gyration    X= 


GYRATION. 


193 


I  126.    GYRATION  OF  A  PARALLELOPIPEDON. 

To  find  the  radius  of  gyration  and  mo-  FIs- 146- 

ment  of  inertia  of  a  parallelopipedon  when 
the  axis  o  o,  of  rotation  passes  through  the 
centre  of  gravity  and  at  right  angles  to 
either  one  of  its  sides. 

The  radius  of  gyration  of  the  dotted 
section  passing  through  the  axis  and  centre 
of  gravity  is  (§  121) 

0.28875  b. 

The  radius  gyration  of  any  other  parallel  section  at  a  distance  I 
from  the  axis  is 

1/P+ (0.28875  b^ 

a  b  =  area  of  one  side  of  the  parallelopipedon  in  the  plane  of  rota- 
tion.    Then  the  differential  moment  inertia  will  be 

ti(a  b) X*  =  0  +  (0.28875  bf[b  til. 
Moment  inertia    a  b  X*  =j[P+ (0.28875  &)*]&  til. 


Integrate  the  moment  inertia  from  I  =  o  to  I  =  \  a,  and  we  have 


Radius  gyration     JT=  -*/ 


Let  the  same  parallelopipedon  revolve  around  an  axis  o',  o'  parallel 
with  o,  o,  and  R  =  radius  of  the  centre  of  gravity.  Then  the  radius 
of  gyration  will  be  (§  123  and  124) 


194 


ELEMENTS  OF  MECHANICS. 


|  127.     GYRATION    OF   A   LATERAL   TRIANGLE    OR    SOLID   WEDGE 
ROTATING   AROUND    ITS   VERTEX. 


Fig.  147. 


h  =  height  and  6  =  linear  base  of  the  triangle. 

a  =  any  breadth  parallel  with  b,  but  at  a  distance  I 

from  o,  o. 

0  =  ^  breadth,  the  area  of  the  triangle. 
The  differential  moment  of  inertia  will  then  be 


a  :  b  =  I  :  h,      of  which      a  =  —  . 
h 

60X2  =  —  91. 
h 


Moment  inertia, 


4A 


Had.  gyration, 


0.70107  A. 


\  128.  GYRATION  OF  A  LATERAL  TRIANGLE  OR  SOLID  WEDGE 
ROTATING  AROUND  AN  AXIS  PASSING  THROUGH  THE  CENTRE 
OF  GRAVITY. 


Fig.  148. 


but 


From  the  preceding  section  we  know  that  when 
the  body  revolves  with  its  vertex  in  the  axis  o'  o',  the 
rad.  gyration  is  0.70107A. 

We  also  know  from  §  123  that  when  r  =  rad. 
gyration  of  a  body  when  rotating  around  ita  centre  of 
gravity, 

X=  T/tf  +  i*.         of  which  r  =  i/X*-I?, 


-3  0.2357A, 


which  is  the  radius  gyration  of  a  triangle  or  wedge  rotating  around 
an  axis  passing  through  the  centre  of  gravity. 


GYRATION. 


195 


|129. 


GYRATION     OF    A    WEDGE    ROTATING    AROUND    AN    AXIS 
PASSING   THROUGH   THE   MIDDLE   OF   ITS   BASE. 


x  - 


Fig-149- 


In  this  case  R  =  \  k,  and  JR?  =  %  h*.     From  the  pre- 
ceding paragraph  we  have  r  =  0.2357A   and  r2  =  -fa  A2. 

x  =  !/£  A*  +  TV  A2  =  Vi  =  0.4082A. 


§130.     CIRCULAR   PLANE   OR   SOLID   CYLINDER. 

Fig.  150  represents   a  circular  plane  or  solid   cylinder  rotating 
around  its  centre. 

Let  A  represent  the  area  of  the  circular  end  of  the      Fig.  wo. 
cylinder,  and  r  =  any  radius  less  than  R.     The  circum- 
ference of  the  radius  r  is  2  x  r,  and  the  differential  area 
ft  A  =  2  JT  r  dr. 

Differential  mom.  in.,     6  A  X*  =  2x  i*8r. 


X* 


but 


then 


r=.   A 

2x1? 


Radius  gyration,        X  =  Jf*  =  - 
\  '2       1 


R 


414 


2" 
-0.7071 


Radius  of  gyration,   X=  0.7071  R. 

§131.     ANNULAR   RING. 

This  example  is  the  same  as  the  foregoing,  except  that  the  formula 


is  integrated  from  r  to  R  instead  of  from  o  to  R. 

s 
X*  = 


2  »  r* 


Fig.  151. 


2^2- 

npw 


Rad.  of  gyrati 


The  radius  of  gyration  of  a  circle  or  a  cylindrical  surface  rotat- 
ing around  its  centre  or  axis  is  equal  to  the  radius  of  the  circle  or 
cylinder. 


196 


ELEMENTS  OF  MECHANICS. 


Fig.  154. 


\  132.     RADIUS  OF  GYRATION  OF  DIFFERENT  FIGURES. 

Fig.  152. 

A  circumference  rotating 
around  its  diameter,  and  a 
circular  plane  or  cylinder 
around  its  centre.  r  =  radius 
of  the  circle. 

Radius  gyration  X=  0.7071  r. 

~W~ 

Fig.  153. 

_l_          A  Circular  Plane  Revolving  around  its  Diameter. 


Radius  gyration     X=  0.5  r. 
r  =  radius  of  the  circle. 

A  Sphere  Revolving  around  its  Diameter. 

Spherical  surface     JT=  0.8165  r. 
A  solid  sphere        X=  0.6324  r. 

A  Cylinder  Rotating  around  one  of  its  Ends. 

v       /4TT3? 

liadius  gyration     JL  =  -*J . 

\      12 


A  Cylinder  Rotating  around  its  Middle. 

Radius  gyration     X" -*/  — — — . 

A  Cone  Revolving  around  its 
Base        X=. 

Vertis     X= 


12  A*+3  I? 
20 


GYRATION. 


197 


Fig.  158. 


A  Cone  Frustum  Rotating  around  its  Base. 


V  / 

JT-J-I 


A  "Wedge,  or  an  Ann  of  a  Wheel. 


=  0.204^/12^ 


A  Cylinder  whose  Centre  Line 

is  parallel  with,  the  axis  of  rotation. 


A  Sphere  Rotating  with  its  Centre 

a  distance  a  from  the  axis. 


Fig.  161. 


A  Ring  of  Square  Section, 

or  a  fly-wheel  of  very  light  arms. 


To  find  the  Common  Radius 

of  gyration  of  a  fly-wheel  with  arms  of 
considerable  weight. 

W=  weight  of  the  ring  of  square  section. 
w  =  weight  of  all  the  arms. 
b  =  breadth  of  the  arms  in  the  direc- 
tion of  rotation  in  feet. 
17* 


198 


ELEMENTS  OF  MECHANICS. 


/6 
~\ 


Fig-  163- 


\  133.     PARALLEL  ARMS  OF  FLY-WHEELS. 

Fig.  163  represents  a  part  of  a  fly-wheel  with  hub  H,  arm  M,  and 
ringZ 

It  is  required  to  find  the  radius 
of  gyration  of  the  arms. 
L  =  length  of  the  parallel  arm. 
A  =  area  of  cross  section  of  the 

arm. 
r  =  radius  of  the  hub. 


1  =  distance  from   the  centre  to 

any  section  A. 

Q  =  cubic  content  of  the  arm. 
y  =  radius  of  gyration  of  the  arm. 
The  momentum  of  radius  of  gyration  of  the  consolidated  element- 
ary sections  A  of  the  arm  will  be 


Q  is  the  variable,  and  not  y. 


f(# 


-i*  L+r  I? 


but     Q  =  AL,    and 


Eadius  gyration    y  •=  |/r*  -f  r  L + %  L9 


FLY-WHEELS.  199 


§  134.     RADIUS  OF  GYRATION  OF  THE  WHOLE  FLY-WHEEL. 

x  =  Radius  of  gyration  of  the  hub,  to  be  calculated  from  §  130. 
0  =  weight  in  pounds  of  the  hub. 

y  =  Radius  gyration  of  the  arms,  to  be  calculated  from  §  133. 
P=  weight  in  pounds  of  all  the  arms. 
2  =  Radius  gyration  of  the   outer  ring,  to   be   calculated   from 

§  130. 

Q  =  weight  in  pounds  of  the  ring. 
X=  Radius  gyration  of  the  whole  wheel. 
W=  weight  in  pounds  of  the  whole  wheel. 

W-O+P+Q. 


Example.  Required  the  radius  of  a  cast-iron  fly-wheel  of  the  follow- 
ing dimensions  ? 

C  r  =  6  inches,  radius  shaft  in  the  hub. 
Hub,     •<   R  =  15  inches,  outside  radius  of  the  hub. 
(  0  =  2798  pounds,  weight  of  the  hub. 

Arms    {  L  =  10'75  feet>  Ien8th  of  the  arms- 

1  P=  10650  pounds,  weight  of  six  arms. 

f  J?2  =  13  feet,  outer  radius  of  the  ring. 
Km&     {  Q  =  36500  pounds. 

Wheel,      W=  49948  pounds,  the  weight  of  the  whole  wheel. 
Radius  gyration  of  the  hub  will  be, 


=  0.952  of  a  foot. 


2 
Radius  gyration  of  the  arms  will  be, 

y  -  y  1.25'+ 1.25  x  10.75 +£  x  10.75*  =  7.32  feet. 
Radius  gyration  .of  the  ring, 

[TTo* 

12.51  feet. 


200  ELEMENTS  OF  MECHANICS. 

Radius    gyration    of   the  whole    system    of   the    fly-wheel  will 
then  be, 


/27 

\ 


Q.9521  +10650x7.32*+  36500  xl2.5  ?  f 


49948 
the  radius  gyration  of  the  fly-wheel. 

This  is  12-  11.217  =0.783  feet,  or  9  inches  less  radius  of  gyration 
than  the  inner  radius  of  the  ring,  which  latter  is  generally  taken  in 
practice.  In  this  case  the  radius  gyration  is  only  0.934  of  the  inner 
radius,  or  0.86  of  the  outer  radius. 

\  135.     FLY-WHEELS. 

In  fly-wheels  of  ordinary  proportions  the  radius  of  gyration  x  can 
be  assumed  in  practice  to  be  the  inner  radius  of  the  ring. 
TF=  weight  of  a  fly-wheel  in  pounds. 
X=°  radius  of  gyration  in  feet. 
n  =  revolutions  per  minute. 
K=  work  in  the  fly-wheel  in  foot-pounds. 
V=  velocity  of  centre  of  gyration  in  feet  per  second. 

WV* 


2         2x32.17' 

/9~  ~P~~, 

of  which         Fs 


^     60     /  2x32.17x60' 

=  5867.16  ' 

This  formula  gives  the  work  of  a  fly-wheel  or  of  any  rotating  body  ; 
that  is  to  say,  it  requires  that  much  work  to  bring  a  body  from  rest 
to  a  rotation  of  n  revolutions  per  minute ;  or  if  the  body  is  rotating 
with  an  angular  velocity  of  n  revolutions  per  minute,  it  will  re- 
generate that  much  work  before  it  is  brought  to  rest. 

1    76.6 


5867.16  ' 

5867.16  K 
X'1  n1      ' 


n    \W 


76.6 


4 


Example  1.  A  fly-wheel  of  TF=2000  pounds,  and  radius  of  gy- 
ration X=  3  feet,  is  to  be  set  in  rotation  by  a  weight  F=  600  pounds, 


FLY-WHEELS.  201 


Fig.  162.     What  angular  velocity  will  the  fly-wheel  obtain  by  the 
weight  falling  50  feet  ? 

The  work       K=  600  x  50  =  3000  foot-pounds. 

hrfi  A         F6AAA 

Revolutions    n  =  —  '—\l  -  =  31.278  per  minute. 
3    \2000 

In  this  case  it  makes  no  difference  what  radius  the  weight  F  is 
acting  upon,  because  the  weight  multiplied  by  the  fall  will  be  the 
work  stored  in  the  fly-wheel  ;  and  the  same  work  will  be  re-utilized 
in  bringing  the  fly-wheel  to  rest. 

Example  2.  A  fly-wheel  weighing  W=  16000  pounds,  and  radius 
of  gyration  X=  6  feet,  is  revolving  at  the  rate  of  n  —  60  revolutions 
per  minute.  Required  the  work  in  the  wheel  ? 

v    16000  x6'x602     Q,Q/10,»    , 
-^=  —  *»/.»  -,f>  -  =  353425  foot-pounds. 
5867.  lo 

The  fly-wheel  can  wind  up  a  weight  of  353425  pounds  one  foot 
high,  or  3534.25  pounds  to  a  height  of  100  feet.  Whatever  height 
divided  into  353425  will  be  the  weight  the  wheel  can  wind  up  to 
that  height. 

§  136.     FLY-WHEELS  IN  REGARD  TO  TIME  AND  SPACE. 

In  the  preceding  paragraph  we  treated  fly-wheels  in  regard  to 
work  and  angular  velocity,  without  regard  to  the  time  and  space 
which  are  constituent  elements  of  work.  We  will  now  treat  on  the 
time  required  for  storing  or  re-storing  the  work  in  a  fly-wheel,  and 
the  space  or  total  number  of  revolutions  in  the  time  required  to  bring 
the  fly-wheel  from  rest  to  a  uniform  rotation  or  from  a  uniform  rota- 
tion to  rest. 

JV=  total  number  of  revolutions  in  the  time  T  seconds. 

The  space  of  uniformly  accelerated  or  retarded  motions  is 


= 
2 

when  V  means  the  uniform  velocity  of  centre  of  gyration. 


2  +  60 


202  ELEMENTS  OF  MECHANICS. 


W  V* 
Wnrlr        7T                   =,  - 

WXW 

2? 
WX* 

5867.16' 
^120mj    2.454317^^* 

5867.16 
^    2.4543  WX*N*          1 

^     T    }               F 

2« 

TP"^                                   ° 

\J\J  -^A.    -i-T-»l     ^^     . 

#-          T        ^                * 

2.4543  X1  N9' 

Example  1.  A  fly-wheel  at  rest  weighing  W=  10000  pounds  is  to 
be  put  into  n  =  80  revolutions  per  minute  in  N=  6  revolutions  of  the 
wheel.  Kequired  the  time  T?  The  radius  of  gyration  being  X=  5 
feet. 

^    10000x5'x80l    oaK,n,  ,.    . 
Work,        K=  —       — — —  =  295504  foot-pounds. 
5867.16 

Time,    T=  1.5666  x  5  x  6^| 10000  _  9.0986  seconds. 

Example  2.  The  fly-wheel  in  the  preceding  example  is  to  be 
stopped  in  T=  5  seconds.  How  many  revolutions  N  will  the  wheel 
make  in  the  action  of  stopping  it  ? 

Revolution,     N=  — ^— J®*®*  =  3.3. 
1.5666  \  10000 


CENTRING. 


203 


£137.     ON   THE   CENTRING   OF   REVOLVING   BODIES. 

This  subject  has  given  much  trouble  to  mechanical  engineers,  for 
the  reason  that  a  body  perfectly  balanced  around  its  axis  of  rotation 
does  not  appear  to  be  balanced  when  set  into  a  high  rotary  velocity. 

Let  the  bodies  IF  and  w  be  connected  by  an  inflexible 
rod  R  r,  and  the  system  perfectly  balanced  on  the  line 
a  b  ;  that  is, 


R  and  r  are  distances  from  the  fulcrum  a  b  to  the  cen- 
tres of  gravity  of  their  respective  bodies. 

Suppose  the  system  to  rotate  in  a  plane  at  right  angles 
to  the  axis  a  b,  then  the  centrifugal  force  of  each  body 
will  be 


29335 


F=J^^_ 2 

2933.5 

But  W  R  =  w  r,  consequently  the  centrifugal  forces  of  the  bodies 
must  be  alike,  and  have  no  tendency  to  disturb  the  equilibrium  in  the 
fulcrum  of  rotation. 

The  work  stored  in  each  body  by  setting  the  system  in  rotation 


wmbe 


3 


586 


in  which  X  and  x  are  the  respective  radii  of  gyration,  but  for  sim- 
plicity in  the  illustration  we  can  (without  much  error)  consider  x  =  R 
and  x  =  ry  and  we  have  the  works, 


5867.16 


_ 
5867.16 


We  have  W  R  =  wr,  butTF-fi?  is  not  equal  to  w  r2,  and  conse- 
quently the  works  stored  in  each  body  are  not  alike.  The  small  body 
w  on  the  greater  radius  r  has  taken  up  more  work  than  has  the 
greater  body  W  on  the  smaller  radius  R  ;  and  it  is  this  difference  of 


204  ELEMENTS  OF  MECHANICS. 

work  which  caused  an  action  in  the  fulcrum  whilst  the  bodies  were 
set  in  rotation.  This  irregular  distribution  of  work  in  revolving 
bodies  has  puzzled  many  good  mechanics. 

Example.    W=  150  pounds  and  JR  =  0.5  feet. 

w  =   20  pounds  and  r  =  3.75  feet. 

n  =  500  revolutions  per  minute. 
Required  the  centrifugal  force?  and  work  stored  in  each  body? 

150  x  0  5  x  5002 
Centrifugal  force      F=  --  —  ^-  -  =  6391.7  pounds  of  each  body. 


The  works  concentrated  in  the  bodies  will  be  respectively 

„     150  x  0.5'  x  500*     . 
A  =  -  -  --  =  1598  foot-pounds. 
5867 


We  see  here  that  the  work  stored  in  the  small  body  w  is  over  seven 
times  greater  than  that  stored  in  the  larger  body  TF;  and  it  is  this 
diiference  which  causes  the  revolving  system  to  work  irregular  in  the 
fulcrum. 

Suppose  the  two  bodies  W  and  w  to  be  cast-iron  balls,  and  find 
their  diameter  and  radii  of  gyration  ?  The  diameters  are  nearly 

Z>  =  0.58  and  d  =  0.207  of  a  foot. 


•r=\^+^=\a5M^ =a533  feet- 

Radius  gyration 

i  »        J*        i  n  90Q5 

=  3.75  feet. 


10 


The  radius  of  gyration  of  the  large  ball  is  0.033  of  a  foot  longer 
than  the  radius  R  ;  but  in  the  small  ball  there  is  no  appreciable  dif- 
ference between  x  and  r. 

The  real  work  stored  in  the  large  ball  will  then  be 

^    150  x  0.533'  x  500' 
JT™  —      -  :  -  =  1815  foot-pounds. 
5867 

or  1815-  1598  =  217  foot-pounds  more  than  in  the  first  calculation. 


TRANSFORMATION  OF  WORK. 


205 


Any  system  of  revolving  bodies  which  is  balanced  when  at  rest 
will  also  be  balanced  in  any  uniform  rotation  ;  but  when  the  rotation 
is  accelerated  or  retarded  work  is  stored  or  re-stored,  which  causes 
an  irregularity  in  the  axis  if  the  several  radii  of  gyration  are  not 
balanced. 


138. 


TRANSFORMATION  OF  STRAIGHT  LINEAR  MOTION   TO 
ROTARY  BY  THE  AID  OF  A  CRANK. 


The  illustration  represents  a  steam-engine,  of  which  the  piston  0 
has  a  straight  linear  motion,  which  is  transformed  into  rotary  by  the 


Fig.  165. 


crank  E.     It  is  supposed  that  the  irregular  motion  of  the  piston  ac- 
commodates itself  to  the  uniform  rotary  motion  of  the  crank. 

Let  F  denote  the  force  acting  constantly  on  the  piston  throughout 
the  stroke ;  then  the  force  acting  on  the  crank  It  in  the  direction  of 
the  tangent  of  the  circle  will  be  F  sin.v.  The  volumes  of  the 
diagrams  represent  the  works  in  the  cylinder  and  crank  respectively 
for  a  single  stroke  or  half  a  revolution  in  the  time  T.  The  work  in 
the  cylinder  is  composed  of  a  constant  force  by  variable  velocity, 
whilst  that  in  the  crank  is  composed  of  a  variable  force  by  constant 
velocity,  but  the  two  works  or  the  products  of  the  three  simple 
FV  Tare  alike  in  both  cases. 

The  variable  velocity  of  the  piston  and  the  variable  force  of  rota- 
tion are  represented  by  the  ordinates  in  the  respective  semicircles. 
The  time  is  represented  in  this  case  by  the  length  of  the  semicircles. 

The  force  and  velocity  in  one  diagram  take  one  another's  place  in 
the  other  diagram. 
18 


206 


ELEMENTS  OF  MECHANICS. 


§139. 


ON    THE    REGULATION    OF    IRREGULAR    WORK    BY 
FLY-WHEELS. 


Fig.  166  represents  a  steam-engine  and  fly-wheel  winding  up  a 
weight,  which  operation  can  be  compared  with  any  uniform  work 
accomplished  by  an  engine.  F°=  constant  force  acting  on  the  piston 
throughout  the  stroke  8. 

Fig.  166. 


Then  F :  W*=  s  :  8  for  each  single  stroke  or  half  a  revolution  of 
the  wheel. 

TT/'ft                             TT  Of  TT7"o                          E*  O 

T-.     ^                -rr^                                     -r-f                 W    S                             -„—           ^    O  rv               "     6                                             -TO 


For  simplicity  in  the  illustration  it  is  supposed  that  the  connecting 
rod  is  infinitely  long,  that  the  steam-pressure  is  constant  throughout 
the  stroke  of  the  piston,  that  all  the  work  of  the  engine  is  transmitted 
through  the  crank-shaft  and  regulated  by  the  fly-wheel  to  a  uniform 
power  in  the  realized  work  of  rotation. 

When  the  force  F  is  constant  throughout  the  stroke  S,  the  work  in 
the  cylinder  can  be  represented  by  the  area  of  a  circle  A  a  b  B  c  d 
for  one  revolution  of  the  engine.  The  area  of  the  semicircle  A  b  e  B 
will  then  represent  the  work  for  a  single  stroke  of  the  piston.  The 
diameter  A  B  is  equal  to  the  stroke  S  of  the  piston,  and  is  sup- 


IRREGULAR    WORK.  207 


posed  to  be  in  the  centre  line  of  the  engine,  so  that  the  crank-pin 
describes  the  circle  and  is  on  the  centre  at  A  and  B. 

When  the  crank  passes  the  centres  A  and  £  the  engine  performs 
no  work  in  the  rotary  motion,  and  when  the  power  in  the  realized 
work  is  constant  in  all  positions  of  the  crank,  the  fly-wheel  must  per- 
form the  full  power  of  the  engine  when  the  crank  passes  the  centres. 
The  power  of  the  engine  varies  as  the  sine  for  the  angle  of  the  crank 
to  the  centre  line  A  &\  and  the  work  performed  in  the  time  of  one 
revolution  of  the  crank  is  represented  by  the  area  of  the  circle  de- 
scribed. 

Draw  on  A  B  the  two  rectangles  A  ef  B  and  B  g  i  A,  equal  to 
the  area  of  the  circle ;  then  the  height  A  =  A  e  represents  the  mean 
velocity  of  the  piston  throughout  the  stroke  S.  Suppose  the  engine 
to  be  running  with  a  uniform  rotation  of  the  fly-wheel,  the  area  of 
the  semicircle  will  then  represent  the  real  work,  and  the  rectangle 
the  mean  work  on  the  piston  for  a  single  stroke. 

It  will  be  seen  that  the  circle  projects  over  the  rectangle  with  a 
segment  a  b,  which  is  the  work  stored  in  the  fly-wheel  during  that 
portion  of  the  stroke,  or  whilst  the  crank-pin  passes  from  a  to  b.  The 
area  of  the  corners  bf  B  and  B  g  c  of  the  rectangles  which  project 
over  the  circle  are  equal  to  the  area  of  the  segment  a  b,  or  the  work 
restored  to  the  rotary  motion  by  the  fly-wheel,  whilst  the  crank  passes 
from  b  via  centre  B  to  c,  and  another  segment  of  work  c  d  is  stored 
in  the  fly-wheel  and  restored  to  the  work  of  rotation  whilst  the  crank 
passes  from  d  via  centre  A  to  a.  The  fly-wheel  is  thus  regulating  the 
irregular  work  in  the  steam  cylinder  to  a  uniform  work  of  rotation. 

3  140.      TO    FIND   THE   IRREGULARITY   OF   ROTATION   OF    A 
FLY-WHEEL. 

The  alternate  storing  and  re-storing  of  work  by  a  fly-wheel  causes 
a  slight  irregularity  of  rotation. 

The   greater   the   fly-wheel   is    and  the  Fig. 

greater  the  velocity  of  rotation,  the  less  will 
the  irregularity  be. 

Suppose  the  crank  and  fly-wheel  to  move 
in  the  direction  indicated  by  the  arrow,  and 
the  rotation  will  be  fastest  when  the  crank- 
pin  passes  b  and  d,  and  slowest  at  a  and  c. 

Let  R  denote  the  radius  of  the  crank. 
The  area  or  work  of  the  semicircle  A  a  b  B 

will  be  -^  -  IF.  The  area  or  mean  work  of  the  rectangle  A  e/B 
will  be  2  R  h. 


208  ELEMENTS  OF  MECHANICS. 

Then        |  rr  I?  =  2  R  h,        of  which        A  =  —  =  0.7853  ^. 

That  is  to  say,  the  mean  velocity  of  the  piston  is  equal  to  0.7853 
of  the  circular  velocity  of  the  crank-pin.  The  fraction  0.7853  is  the 
sine  for  the  angle  v  of  the  crank  with  the  centre  line  of  the  engine 
when  the  rotary  motion  is  a  maximum  at  b  and  d,  and  minimum  at 
a  and  c.  The  angle  v  =  51°  45'. 

The  work  k  stored  and  re-stored  alternately  by  the  fly-wheel  or 
the  area  of  the  segment  a  b  will  be 


-  1?  sin.v  cos.v. 


180 
v  =  51.75°.  sin.v  =  0.  7853.  cos.v  =  0.61909. 

k  -  8'14  X  38"25^  _  Q.7853  x  0.61909  ^  =  0.18109^. 

This  is  the  area  of  the  segment  a,  6  expressed  in  a  fraction  of  a 
square  radius  ;  but  we  want  the  segment  of  work  to  be  expressed  in 
a  fraction  of  the  semicircle,  or  of  the  work  of  a  single  stroke  of  the 
piston. 

The  area  of  a  semicircle  «=  \K  H?. 

Call  .F#=the  work  of  a  single  stroke, 


Then    ^ 


That  is  to  say,  the  irregular  work  stored  and  re-stored  by  a  fly- 
wheel is  equal  to  0.1153  of  the  work  of  a  single  stroke  of  the  piston. 
As  the  same  proportion  is  constant  for  every  stroke  of  the  engine,  it 
follows  that  the  work  of  regulation  by  a  fly-wheel  is  always  0.1153 
of  the  work  of  the  engine,  when  the  steam-cylinder  is  double  acting 
and  the  realized  work  is  uniform. 


IRREGULAR    WORK. 


209 


ON    IRREGULARITY   OF   ROTATION    OF   A   FLY-WHEEL. 

Notation  of  Letters  (repeated). 


Fig.  168. 


B 


W=  weight  of  fly-wheel  in  pounds. 
x  =  radius  of  gyration  in  feet. 
n  =  number  of  revolutions  per  minute. 
K=  work  in  foot-pounds  in  the  fly-wheel. 
/=  irregularity  in  fraction  of  the  revo- 
lutions n,  or  of  the  mean  velocity 
of  rotation. 

The   mean   velocity  of   the   fly-wheel  is 
when  the  crank  passes  the  centres  at  A  and 

£,  the  greatest  velocity  at  b  and  d,  and  the  slowest  at  a  and  c. 
Whilst  the  crank-pin  passes  from  A  to  a  the  work  represented  by 
the  projecting  corner  A  e  a  is  given  out  from  the  fly-wheel,  by  which 
the  velocity  is  reduced. 

The  work  of  the  corner     A  ea  =  £x0.1153  F8=  0.0576  F8. 

W  V2  <n- 

The  work  in  the  fly-wheel  is     K=  -     -  --. 

5867.16 


Mean  revolutions     n  = 


76.6 


Revolutions  or  angular  velocity  at  a, 

,     76.6    \K-  0.0576  FS 

n-—\~ 


W 


Irregularity    /= 


/.I-^.M™.    ! 


338  FS 


18.4 


18.4 


i-V1-  r^ 

FS 


FS 


n    \F[1-(1-/)'] 

Example.  A  fly-wheel  of  TF"=  12000  pounds,  and  radius  of  gyration 
x  =  7  feet,  is  to  make  n  =  48  revolutions  per  minute,  with  an  engine 
of  S=  4  feet  stroke  and  steam-pressure  on  the  piston  F=  12750  pounds. 
Required  the  irregularity  of  rotation  of  the  fly-wheel? 


/=!- 


^388x12750x4 
12000  x72x  48s 
o 


210 


ELEMENTS  OF  MECHANICS. 


This  very  small  irregularity  could  not  be  perceived  without  a 
delicate  dynamometer,  and  the  fly-wheel  is  sufficiently  large. 

The  irregularity  should  never  exceed  0.1,  and  in  ordinary  practice 
/  should  be  from  0.01  to  0.05. 

When  /=  1  the  fly-wheel  cannot  carry  the  engine  over  the  centre. 
/=!,  when  338  F3=  Wx*n\ 

$142.     FLY-WHEEL   FOR   A   STAMP-MILL. 


Notation  of  Letters. 


Fig.  169. 


W=  weight  of  fly-wheel  in  pounds. 
w  =  weight  of  stamp  in  pounds. 
8=  vertical  height  in  feet  which 

the  stamp  is  lifted. 
n  =  number  of  blows  per  minute. 
e  =  angle  which  the  cam  rotates 

whilst  lifting  the  stamp. 
IP  =  horse-power  in  operation. 
The  work  done  by  each  blow  will  be 

K=wS.        .     1 
Per  minute        K=w  Sn.     .     2 


Horse-power 


33000 


This  horse-power  is  supposed  to  be  constant  in  driving  the  shaft 
and  fly-wheel,  and  a  part  of  the  work  of  that  power  is  stored  in  the 
fly-wheel  whilst  the  cam  rotates  freely  without  lifting  the  stamp. 

The  work  stored  in  the  fly-wheel  between  each  lift  is 

0°-e\ 


360° 


which  work  is  re-utilized  in  lifting  the  stamp. 

N=  rate  of  revolutions  per  minute  at  the  moment  the  cam  com- 
mences to  lift  the  stamp,  when  the  work  stored  in  the  fly- 
wheel is 


5867.16  ' 


STAMP-MILL.  211 


u  =  rate  of  revolutions  at  the  moment  the  cam  drops  the  stamp, 
when  the  work  stored  in  the  fly-wheel  is 

. 

•' 


58676- 


360         5867.16 
5867. 


r.16  W/SV360  -e\ 

iF*~~\~36o~; 


.,    5867.16  w^,^    „! 

•          I 


360 
The  irregularity  of  rotation  of  the  fly-wheel  will  then  be 


_  5867.  -  0 

"1"  -~-    '        '    J 


Example.  A  stamp  weighing  w  =  1200  pounds  is  to  be  lifted  $  =  3 
feet  n  =  48  times  per  minute.  The  weight  W=  9720  pounds  and 
radius  gyration  x  =  5.25  feet ;  the  cam  moves  e  =  120°  in  lifting  the 
stamp.  Required  the  irregularity /of  the  rotation  of  the  wheel? 


5867.16 


This  irregularity  is  very  small,  and  the  fly-wheel  is  consequently 
sufficiently  large,  and  could  be  made  a  little  smaller. 

The  work  consumed  direct  from  the  motive-power  in  raising  the 
stamp  is 


and  the  work  taken  out  from  the  fly-wheel  is 


The  total  work  consumed  in  raising  the  stamp  is 


X  =  w  s(S60~e+e\=  w  S,  the  realized  work.  .  11 

\      36°     / 


212 


FLY-WHEELS. 


ELEMENTS  OF  FLY-WHEELS. 

Outside 
diameter 
of  wheel. 

Square 
section 
of  ring. 

Radius 
of 
gyration. 

Weight 
of 
wheel. 

Work 
K=Ctf 

Mass 

M 

feet. 

inches. 

x  in  feet. 

TTinlbs. 

coef.  C 

rnatts. 

2 

2 

0.82 

120 

0.0138 

3.7202 

2 

2.5 

0.80 

180 

0.0196 

5.5853 

2 

3 

0.76 

270 

0.0266 

8.3930 

2.5 

2.5 

1.03 

234 

0.0424 

7.2739 

2.5 

3 

1.00 

336 

0.0625 

10.444 

2.5 

3.5 

0.95 

460 

0.0710 

14.300 

3 

2.5 

1.30 

281 

0.081 

87.347 

3 

3 

1.20 

405 

0.100 

12.589 

3 

3.5 

1.00 

550 

0.0937 

17.095 

3.5 

3 

1.50 

472 

0.180 

14.673 

3.5 

3.5 

1.30 

640 

0.184 

19.895 

3.5 

4 

1.10 

840 

0.174 

26.112 

4 

3 

1.75 

540 

0.282 

16.784 

4 

4 

1.42 

960 

0.330 

29.842 

5 

3 

2.25 

675 

0.582 

20.973 

5 

4 

2.17 

1,200 

0.960 

37.202 

6 

4 

2.66 

1,440 

1.750 

44.763 

6 

5 

2.60 

2,250 

2.600 

81.721 

8 

5 

3.60 

3,000 

6.63 

93.255 

8 

6 

3.50 

4,320 

9.03 

134.29 

10 

6 

4.50 

5,400 

18.65 

167.84 

10 

8 

4.20 

9,600 

28.9 

298.42 

12 

6 

5.50 

6,480 

33.4 

201.43 

12 

9 

5.25 

9,720 

45.8 

302.15 

16 

9 

7.25 

12,960 

116.5 

402.86 

16 

12 

7.00 

34,560 

289 

1074.3 

20 

9 

9.25 

24,300 

355 

755.37 

20 

12 

9.00 

43,200 

598 

1342.9 

25 

12 

11.50 

64,800 

1460 

2014.3 

25 

15 

11.25 

84,300 

1820 

2620.5 

30 

12 

14.00 

65,800 

2200 

2045.4 

30 

15 

13.75 

100,250 

3230 

3116.3 

36 

15 

16.75 

120,500 

5740 

3735.7 

36 

18 

16.5       1     170,000 

7500 

5274.4 

The  weight  of  cast-iron  fly-wheels  is  closely  approximated  by  the 
following  rule : 

Multiply  the  cross-section  of  the  ring  in  square  inches  by  the  out- 
side diameter  of  the  wheel  in  feet,  and  the  product  by  15,  which  will 
be  the  weight  in  pounds  of  the  fly-wheel,  including  arms  and  centre. 

The  square  of  the  number  of  revolutions  per  minute  multiplied  by 
the  coefficient  in  the  column  C  will  be  the  work  in  the  wheel. 


FORGE-HAMMER.  213 


\  143.     FORGE-HAMMER  WORKED  BY  CAMS. 

Fig.  170. 
(\ 

The  illustration  presents  a  forge - 
hammer  worked  by  four  cams. 
The  fly-wheel  is  on  the  shaft  c.  V_-X  .=«. 


Notation  of  Letters. 

W=  weight  of  fly-wheel  in  pounds. 

w  =  weight  of  the  hammer  in  pounds,  including  its  whole  moving 

system. 

tS  =  vertical  space  in  feet  which  the  centre  of  gravity  of  the  ham- 
mer is  lifted. 

n  =  number  of  revolutions  of  the  fly-wheel  per  minute. 
N=  number  of  lifting  cams  in  the  circle. 
E=  angle  in  degrees  between  each  cam. 
e  =  angle  which  the  cams  rotate  whilst  lifting  the  hammer. 
IP  =  horse-power  in  operation. 
x  =  radius  of  gyration  of  the  fly-wheel. 
/=  irregularity  of  rotation  of  the  fly-wheel  between  each  blow  of 

the  hammer. 
The  work  done  in  each  blow  of  the  hammer  is 

K=WS.        .  .        :      .  .    12 

Per  minute         K=wSNn.      .        .        .13 

.     wSNn 

Horse-power      BP  =  —     — .     .    '    .         .14 
33.000 

This  is  the  horse-power  required  to  drive  the  hammer  in  making 
Nn  blows  per  minute. 
The  work  stored  in  the  fly-wheel  between  each  blow  is 

.  15 


which  work  is  re-utilized  in  lifting  the  hammer. 
The  total  work  stored  in  the  fly-wheel  is 


214 


ELEMENTS  OF  MECHANICS. 


The  irregularity  of  work  between  each  lift  of  the  hammer  will 
then  be 


E 


17 


The  irregularity  of  rotation  of  the  fly-wheel  between  each  blow  of 
the  hammer  will  be 


Xn 


is 


Example  1.  A  hammer  like  that  represented  by  the  illustration, 
weighing  w  =  12000  pounds,  is  worked  with  N=  4  cams  making  n  =  10 
revolutions  per  minute,  and  the  centre  of  gravity  of  the  hammer  is 
lifted  $=  1  foot  each  blow.  Four  cams  make  the  angle  E=9Q° 
and  the  angle  e  =  30°.  The  weight  of  the  fly-wheel  is  17=80000 
pounds.  Required  the  irregularity  of  the  fly-wheel  for  each  blow  of 
the  hammer  ?  Radius  of  gyration  of  the  fly-wheel  is  X=  12  feet. 

76.6      /12000xl/90-30\ 
•'~12xlO\    80000    (     90     / 

the  irregularity  of  the  fly-wheel,  which  is  rather  too  much. 

It  is  supposed  in  this  example  that  the  fly-wheel  is  on  the  same 
shaft  as  the  cams,  but  if  placed  on  another  shaft  and  geared  to  make 
more  revolutions  than  the  cam-shaft,  the  irregularity  would  be  reduced 
in  proportion  to  the  gearing,  as  the  formula  will  give  by  inserting  for 
n  whatever  revolutions  the  fly-wheel  makes  per  minute. 


144.     ILLUSTRATION   OF  IRREGULAR   ROTATION  OF  A   STEAM- 
ENGINE   CRANK   WITH    FLY-WHEEL. 

Let  A  B  represent  the  centre  line  of  a 
steam-engine  and  CD  at  right  angles  thereto. 
^ =  rac^us  °f  the  crank,  which  will  be  on 
fae  cen{;re  afc  £  an(j  gt  an(j  at  rignt  angles 

to  the  engine  in  the  points  G  and  D. 

Let  the  circle  AC  B  D  described  by  the 
crank-pin  represent  the  mean  angular  ve- 
locity, and  the  ellipse  abed  the  irregularity 
of  rotation  ;  that  is,  the  distance  from  the 
centre  to  the  periphery  of  the  ellipse  repre- 
sents the  velocity  of  rotation  in  that  position  of  the  crank,  which  is 
slowest  at  a  and  c,  and  fastest  at  b  and  d. 

The  greatest  distance  between  the  circle  and  ellipse  at  the  points 
ab  c  and  o?is  the  irregularity/  corresponding  to  the  preceding  formulas. 


SIZE  OF  FLY-WHEELS.  215 

It  is  supposed  here  that  the  steam-pressure  on  the  piston  is  con- 
stant throughout  the  stroke,  and  that  the  realized  work  is  uniform  in 
the  rotation. 

When  the  steam  is  expanded  half-stroke  the  ellipse  will  be  nearer 
the  circle. 

§  146.     TO  APPROXIMATE  THE  SIZE  AND  WEIGHT  OF  A  FLY-WHEEL 

for    one   double-acting   steam-cylinder  when    the    realized    work    is 
uniform. 

Assume  that  the  ring  of  the  fly-wheel  is  of  a  square  section,  that 
the  radius  of  gyration  is  equal  to  the  inner  radius  of  the  ring,  and 
that  the  weight  of  all  the  arms  and  of  the  hub  is  equal  to  half  the 
weight  of  the  ring.  Assume  also  the  side  of  the  square  section 
of  the  ring  to  be  one  inch  for  each  foot  of  the  radius  of  gyration,  it 
follows  that  the  weight  of  the  fly-wheel  will  be 


J?  x  2  *x  1.04166  x  450x1} 
~~ 


M8FS 


We  may  limit  the  irregularity  of  the  wheel  to  f=  0.025  ; 
then  [1- (1-0.025)']  =  0.05, 

.     33SFS  .    338  FS 

and     30JF5=  —      — ,        and     X5-—      — . 
0.05  tf  1.5  n* 


Example.  The  cylinder  of  a  steam-engine  is  24  inches  diameter 
by  S=  4  feet  stroke.  Steam-pressure  50  pounds  to  the  square  inch  ; 
area  of  piston  452.39  square  inches  x  50  pounds  =  22619.5  pounds  =  F 
the  force  on  the  piston.  The  engine  to  make  w  =  60  revolutions  per 
minute.  Required  the  size  and  weight  of  the  fly-wheel  ? 


Radius  gyration,      X=  3  —  -  5.716, 

say  6  feet  the  inside  radius  of  the  ring. 

The  side  of  the  section  of  the  ring  should  be  6  inches,  and  the  out- 
side diameter  of  the  wheel  will  be  2x6  +  1  =  13  feet. 

Weight  of  the  fly-wheel  =  30  x  63  =  6480  pounds. 


216 


ELEMENTS  OF  MECHANICS. 


Fig.  172. 


FLY-WHEEL    FOR    TWO    ENGINES    CONNECTED   AT   RIGHT 
ANGLES. 

When  two  engines  are  connected  by  cranks  at  right  angles  on  a 
shaft,  each  cylinder  conveys  a  work  like  that  represented  by  the  pre- 
ceding diagrams. 

The  one  cylinder  alone  produces  this  dia- 
gram of  work,  Fig.  172,  as  before  described. 
The  centre-line  being  in  the  direction  A  B. 

The  other  engine,  connected  by  a  crank  at 
lJJ  right  angles  to  that  of  the  former,  produces 
a  diagram  like  Fig.  173.  The  centre-line 
A  B  of  the  engine  being  at  right  angles  to 
that  of  the  former. 

Now  place  the  two  diagrams  on  the  top 
of  one  another,  and  we  have  an  illustration, 
Fig.  174,  of  the  joint  action  of  the  two  engines. 

It  will  be  seen  that  the  segments  of  work 
which  were  stored  and  re-stored  by  the  fly- 
wheel for  each  alternate  stroke  of  one  engine 
alone,  are  now  utilized  direct  to  the  rotary 
motion;  and  the  small  irregularity  of  rota- 
tion is  represented  by  the  black  corners  out- 
side of  the  circle ;  which  shows  that  the 
crank  will  move  a  little  faster  when  at 
angles  of  45°  on  each  side  from  the  centre- 
line, and  slowest  when  passing  the  centres, 
supposing  that  no  fly-wheel  is  employed; 
but  with  a  fly-wheel  the  rotation  will  be 
fastest  when  the  crank  passes  the  black 
points,  which  are  at  angles  of  51°  45',  and 
slowest  when  the  cranks  pass  5°  15'  from 
their  respective  centre-lines. 


|147.     WHEN   FLY-WHEELS   ARE   AND   ARE    NOT   NECESSARY. 

The  duty  of  a  fly-wheel  is  to  regulate  irregular  work,  which  may 
be  either  or  both  the  primitive  or  realized  works.  We  have  heretofore 
treated  the  fly-wheel  as  regulating  the  irregular  primitive  work  in 
the  steam-cylinder,  to  regular  realized  work  of  rotation. 

In  the  case  of  a  steam-engine  lifting  a  hammer  or  a  stamp  by  a 
cam,  both  the  primitive  and  realized  works  are  irregular. 


MOMENTUM-CUR  VE. 


217 


When  a  water-wheel  is  lifting  a  hammer  or  a  stamp  by  a  cam,  the 
primitive  work  is  regular  and  the  realized  irregular. 

In  the  case  of  a  rolling-mill  worked  by  a  steam-engine,  the  irreg- 
ularity of  the  primitive  work  is  so  small  compared  with  that  of  the 
realized  work  in  the  mill,  that  the  work  of  the  fly-wheel  must  be  pro- 
portionate entirely  to  that  of  the  mill. 

In  the  cases  of  pumping-  and  blowing-engines,  where  the  primitive 
work  is  applied  direct  to  the  realized  work,  there  is  no  work  per- 
formed at  the  ends  of  the  stroke,  except  that  of  moving  valves  and  in 
overcoming  friction ;  but  when  two  such  engines  are  connected  by 
cranks  at  right  angles  on  a  shaft,  the  one  engine  will  move  the  valves 
for  the  other  one,  and  no  fly-wheel  is  there  required ;  in  fact,  such 
arrangements  will  work  much  better  without  a  fly-wheel,  like  in 
marine  engines. 

In  upright  blowing-engines  the  heavy  moving  parts  require  a  fly- 
wheel for  regulating  the  motion,  but  if  the  weight  of  the  moving  parts 
are  balanced,  which  can  be  done  by  steam,  the  fly-wheel  would  be  of 
no  utility,  when  two  engines  are  connected  at  right  angles. 

In  horizontal  blowing-  and  pumping-engines  the  moving  parts  are 
well  balanced  and  require  no  fly-wheel. 

When  the  steam  is  expanded  in  the  cylinders,  the  one  engine  will 
work  with  full  steam  whilst  that  of  the  other  is  expanding,  and  they 
thus  help  one  another  alternately  without  fly-wheel. 

|  148.     ROTARY  MOMENTUM  IN  CRANK  MOTION. 

Let  the  circle  A  a  B  b  represent  that  described  by  a  crank  of 
radius  R,  and  A  B  being  the  centre 
line  of  the  steam-engine.  For  sim- 
plicity in  the  illustration  we  will 
suppose  the  connecting-rod  to  be  in- 
finitely long,  when  the  force  F  will 
act  on  the  crank-pin  in  a  direction 
parallel  to  that  of  the  centre  line 
A  B,  and  the  static  momentum  in 
any  position  of  the  crank  will  be  the 
product  of  the  force  jPand  the  sine  (TO) 
for  the  angle  of  the  crank  to  the  line 
A  B. 

Set  off  the  sine  m  from  the  centre  on  its  radius  for  any  number  of 

positions  of  the  crank,  and  join  the  outer  ends,  which  forms  the  curve 

of  rotary  momentum.     When  the  connecting-rod  is  infinitely  long, 

like  in  the  case  of  slot-crank  motion,  the  curve  of  rotary  momentum 

19 


218 


ELEMENTS  OF  MECHANICS. 


will  be  two  circles  inscribed  in  the  circle  of  the  crank,  as  shown  by  the 
illustration. 


1 149.     ENGINE  WITH  CONNECTING-ROD  OF  DEFINITE  LENGTH. 

When  the  connecting-rod  is  of  definite  length  the  momentum  of 
rotation  will  not  be  as  the  sine  of  the  angle  of  the  crank,  but  as  the 
rightangular  distance  from  the  centre  to  the  direction  of  the  con- 
necting-rod. 

1= length  of  the  connect- 
ing-rod. 

JR  =  radius  of  the  crank. 

- 176-  v  =  angle  of  the  crank. 

x  =  angle  of  the  connect- 
ing-rod. 
F=  force  on  the  piston  in 

pounds. 

P  —  pressure  in  the  guides. 
d=  distance  from  centre 
of  crosshead  to  cen- 
tre of  crank-shaft. 


I :  R  =  sin.v  :  sin.x. 


12  sin.v 


. 
d= 


Isin.y     I  sin.(l8Q  —  x  -  v) 
-  -     -  -  - 


sin.v.  sin.v. 

The  lever  m  for  the  momentum  of  the  force  .Fwill  then  be 

,    .         I  sin.(18Q  -  x  -  v)sinjc 
m  =  d  sin.x  = * ; . 


The  lever  m  can  thus  be  calculated  or  constructed  graphically  for 
different  positions  of  the  crank.  The  lever  m  is  set  off  from  the  centre 
on  the  corresponding  position  of  the  crank  which  forms  the  curve  of 
rotary  momentum. 

The  pressure  in  the  guides  will  be 


P^Ftan.x. 


MOMENTUM-  C  UR  VES. 


219 


Fig.  178. 


§150.     MOMENTUM-CURVE   FOR   A   SINGLE   ENGINE 
with  connecting-rod  twice  the  stroke. 

Fig.  177  shows  the  curve  of  rotary  Fig.  ITT. 

momentum  when  the  length  of  the  con- 
necting-rod is  twice  that  of  the  stroke. 
The  shorter  the  connecting-rod  is  the 
more  do  the  curves  lean  toward  the 
engine.  The  outer  curves  show  the 
rotary  momentum  when  the  steam- 
pressure  is  constant  throughout  the 
stroke,  and  the  inner  curves  that  when 
the  steam  is  expanded  f  of  the  stroke. 

§151.     MOMENTUM-CURVE    FOR    TWO    ENGINES    WORKING 
at  right  angles  and  connecting-rods  infinitely  long. 

Curves  of  rotary  momentum  for  two 
engines  working  at  right  angles  on  one 
common  crank,  which  is  the  same  as 
when  two  engines  are  working  on  cranks 
at  right  angles  on  a  common  shaft.  The 
connecting-rods  are  supposed  to  be  in- 
finitely long,  like  when  a  crank-pin  works 
in  a  slot.  The  circle  A  a  B  b  is  that 
described  by  the  crank-pin  around  the 
centre  C.  The  momentums  for  each  en- 
gine are  added  for  each  position  of  the 
crank,  for  which  the  curves  of  rotary  momentum  extend  outside 
of  the  circle  of  rotation.  The  outer  curves  are  for  full  steam,  and 
the  inner  ones  for  %  expansion. 

|  152.     MOMENTUM-CURVE    FOR    TWO    ENGINES    WORKING 
at  right  angles  and  connecting-rods  twice  the  stroke. 

Fig.  179  shows  the  rotary  momentum  Fig.  ITS. 

for  two  engines  working  at  right  angles 
on  one  crank,  and  the  connecting-rod 
being  twice  the  stroke. 

The  inner  curve  shows  the  rotary  mo- 
mentum when  the  steam  is  cut  off  at 
one-third  the  stroke,  or  with  two-thirds 
expansion. 

The  curves  of  rotarv  momentum  serve 


220  ELEMENTS  OF  MECHANICS. 

to  show  the  distribution  of  power  and  work  in  the  circle  described  by 
the  crank-pin. 

In  the  illustrations  of  rotary  momentum  it  is  supposed  that  all  the 
primitive  work  is  transmitted  through  the  cranks  and  shaft,  which  is 
most  generally  the  case  in  steam-engines ;  but  when  the  primitive 
work  is  transmitted  directly  to  the  realized  work,  like  in  blowing 
and  pumping-engines,  where  the  steam-piston  is  connected  directly 
by  a  rod  to  the  pumping  piston,  only  a  small  portion  of  the  work  is 
transmitted  alternately  through  the  crank  and  shaft  as  rotary 
momentum. 

|153.     CHANGING   DIRECTION   OF   A   MOVING   MASS. 

A  free  moving  mass  cannot  within  itself  change  its  direction  of 
straight  linear  motion. 

The  direction  of  a  moving  mass  can  be  changed  only  by  applying 
a  force  in  a  different  direction. 

A  mass  M  is  moving  with  a  uniform  velocity  in  the  direction  a  c, 
but  at  a  a  constant  force  F  is  applied  at  right  angles  to  a  c,  so  that 
the  force  would  move  the  mass  from  a  to  b  in  the  same  time  as  the 
mass  of  itself  would  move  from  a  to  c. 


M 


The  resultant  of  the  two*  motions  will  carry  tbe  mass  from  a  to  c? 
in  that  same  time.  As  the  force  F  is  constant,  the  motion  in  the  direc- 
tion a  b  or  c  d  will  not  be  uniform,  but  accelerated  in  accordance  with 
the  laws  of  dynamics  of  matter — namely,  that  the  spaces  passed 
through  by  a  mass  acted  upon  by  a  constant  force  are  as  the  square 
of  the  times  of  action.  Therefore,  the  mass  M  will  not  move  in  a 
straight  line  from  a  to  d,  but  in  the  course  a,  e,  d,  which  is  a  pa- 
rabola. 

A  jet  of  water  flowing  from  a  vertical  orifice  shows  this  parabolic 
course,  in  which  case  the  force  ^is  the  force  of  gravity,  the  mass  J/ 
is  the  water  which  is  set  in  motion  by  the  pressure  from  behind  the 
orifice. 


HOLE  BY  A   MUSKET- BALL.  221 


\  154.     FIRING   A   MUSKET-BALL  THROUGH  A  DOOR. 

Fig.  181  represents  a  horizontal  section  of  a  door  movable  on  hinges. 

A  musket-ball  V  is  fired  so  as  to  penetrate  the  door  at  a.  It  has 
been  found  by  experiment  that  it  requires  a  force  of  100  pounds  to 
force  the  ball  slowly  through  the  door  one  inch  thick. 

The  work  of  penetration  is  then —  =  8.33  Fig' : 

foot-pounds. 

W=  0.071  of  a  pound,  the  weight  of  the  ball. 
V=  1000  feet  per  second,  velocity  of  the  ball. 
v  =  velocity  after  penetration.  ~:^-^~> 

The  work  .ZT  stored  in  the  ball  before  striking  is 
WV*    0.071  xlOOO* 


After  penetration  the  work  is  £  =  1103.6-8.3  =  1095  foot-pounds. 
Velocity  v  after  penetration  will  be 

~Wq        11095x32.17     ,._,.,    , 
-\  F     \— a07T~   =™.4  feet  per^cond.    ^ 

The  loss  of  velocity  by  penetration  is  295.6  feet  per  second. 
The  mean  velocity  through  the  door  was  852  feet  per  second. 
The  time  of  penetration  was 

T= =  0.0000097844  of  a  second. 

The  door  is  of  white  pine,  3  feet  wide  by  6  feet  high,  which  will 
weigh  52  pounds.  Suppose  the  door  to  move  freely  without  friction 
in  the  hinges  and  without  resistance  of  air.  The  ball  penetrates  the 
door  in  the  centre,  or  at  r  =  1.5  feet  from  the  line  of  the  hinges. 

Radius  of  gyration  of  the  door  is  X=  3x0.5775  =  1.7325  feet. 

In  accordance  with  the  assumed  conditions  the  door  will  be  set  into 
rotation  by  the  penetration  of  the  ball. 

From  §  112  we  have  the  number  of  revolutions  per  minute 
=  60  Fr9  T  =  60  x  100  x  1-5  x  0.0000097844  =  065851 
~  2  TT  W  X"1  6.28  x  0.071  x  1.73252 

of  one  revolution,  or  it  would  require  15.1  minutes  for  the  door  to 
make  only  one  revolution,  which  motion  would  hardly  be  perceptible, 
but  the  resistance  of  air  and  friction  in  the  hinges  would  keep  the 
door  almost  stationary. 


222  ELEMENTS  OF  MECHANICS. 


§  155.  CENTRIFUGAL,  AND  CENTRIPETAL, 
FORCES. 

A  mass  M  moving  with  a  velocity  V  in  the  circle  of  radius  R 
around  the  centre  0  will  have  a  tendency  to  move  outward  from  the 
centre,  for  which  a  force  equal  to  that  tendency 
must  be  applied  on  the  mass  from  the  centre  C,  to 
keep  it  revolving  in  the  circle. 

The  tendency  of  the  mass  to  move  outward  is 
called  centrifugal  force,  and   the   force  resisting 
that  tendency  is  called  centripetal  force.     These 
two  forces  are  equal  and  in  opposite  directions ; 
neither  one  of  them  can  exist  without  the  other, 
in  fact,  their  distinction  is  only  action  and  re- 
action, the  only  condition  under  which  force  can  be   conceived  or 
realized. 

Tie  a  stone  at  the  end  of  a  string  and  swing  it  round  in  the  air, 
and  a  force  is  felt  in  the  hand  which  you  may  call  centrifugal  or 
centripetal  as  you  please.  The  tension  of  the  string  is  the  centri- 
fugal force,  and  the  reaction  by  the  hand  is  the  centripetal  force. 

§156.     THEORY  OF   CENTRIFUGAL   FORCE. 

Let  a  mass  M,  Fig.  180,  move  with  a  uniform  velocity  V  in  the 
direction  a,  b  until  it  arrives  at  a,  where  a  constant  force  F  is  applied 
on  it  at  right  angles  to  the  motion,  which  causes  the  body  to  deviate 
from  the  straight  line  a,  b ;  but  as  that  force  acts  at  right  angles  to 
the  motion,  it  has  no  effect  on  the  velocity  V  of  the  mass.  Suppose 
the  direction  of  the  force  F  to  vary  with  the  deviation  of  the  motion 
from  the  straight  line,  so  as  to  always  be  at  right  angles  to  the  di- 
rection of  motion ;  it  follows  that  the  velocity  of  the  mass  will  be 
constant.  When  the  force  F  is  so  balanced  as  to  cause  the  mass  to 
move  in  a  circle  of  radius  JR  around  the  centre  C,  the  mass  will  then 
return  to  the  same  point  a  where  the  constant  force  was  first  applied. 
If  the  force  F  ceases  to  act  at  the  moment  it  returns  to  a,  the  mass 
will  continue  with  the  same  velocity  in  its  original  course  a,  b,  which 
is  the  tangent  to  the  circle.  Therefore,  when  the  centripetal  force  of 
a  revolving  body  ceases  to  act,  the  body  will  fly  off  in  the  direction 
of  the  tangent  to  that  point  of  the  circle  where  it  was  let  loose. 

Although  the  velocity  of  the  mass  was  not  changed  in  its  circular 
motion,  the  force  F  actually  stopped  the  mass  at  c,  giving  it  a  back- 
ward motion  at  d,  stopped  it  at  e  and  returned  it  into  a  forward 
motion  again  with  the  same  velocity  V  at  a,  all  in  reference  to  the 


CENTRIFUGAL  FORCE. 


223 


direction  a,  b ;  which  operation  was  accomplished  in  the  space  of  the 
diameter  of  the  circle. 

It  follows  that  F  is  equal  to  a  force  which  would  give  the  mass  M 
a  velocity  Fin  a  space  equal  to  the  diameter  of  the  circle,  and  which 
is  the  centrifugal  force  of  the  mass. 


We  have     F:  M  =  F:  T,     F 

VT 

,    an 

M  V 


Diameter    2  E  = 


MV 


a  •  aad  T-v 


Centrifugal  force     F= 


F=  force  in  pounds,  JfeT=mass  expressed  in  matts.,  V=  velocity  in 
feet  per  second  of  the  mass  in  the  circle  of  radius  R  in  feet. 

{157.  PROOF  OF  CENTRIFUGAL  FORCE. 

Fig.  184. 


A  body  or  mass  M,  Fig.  18,£  moving  with  a  uniform  velocity  V  in 
the  direction  A  G,  when  arriving  at  A  a  force  F  is  applied  at  right 
angles  to  the  motion,  which  changes  the  course  of  the  body.  Let  the 
direction  of  the  force  F  be  so  changed  as  to  always  be  at  right  angles 
to  the  motion  of  the  mass  M,  and  it  will  have  no  effect  upon  the 
velocity  V  in  the  path  of  motion.  The  force  F  can  be  of  such  mag- 
nitude as  to  cause  the  body  to  describe  a  circle  A,  £,  C,  D,  F  of 
radius  _K.  Having  given  the  mass  M,  velocity  V  and  radius  E,  the 
problem  is  to  find  the  magnitude  of  the  force  F. 


224  ELEMENTS  OF  MECHANICS. 

When  the  body  has  arrived  at  B,  the  force  F  acting  toward  the 
centre  H  can  be  resolved  into  two  forces  a  and  b,  of  which  a  acts  to 
stop  the  motion  in  the  direction  parallel  to  A,  G,  and  b  acts  to  set  the 
body  in  motion  in  the  direction  parallel  to  A,  H. 

a  =  F  sin.x.     and     b  =  F  cos.x. 

In  the  position  C we  have     c  =  Fsin.<f>,     and    d=Fcos.<p. 
At  D,     e  =  F  sin,z,      and    f=°F  cos.z. 

When  the  body  arrives  at  E,  its  motion  in  the  direction  parallel  to 
A  G  has  been  stopped  by  the  forces  a,  c,  e,  and  finally  F.  If  the 
variable  force  which  stopped  the  body  in  the  direction  A  G  had  been 
applied  opposite  the  motion  at  A,  it  would  have  stopped  the  body  at  G. 
Set  off  the  forces  a,  c,  e  and  ^at  right  angles  to  A  G  in  the  respective 
positions  of  the  body.  Join  these  forces  with  A  and  in,  which  will 
be  a  straight  line.  Then  the  area  of  the  triangle  A,  m,  G  represents 
the 

•Tl     T) 

Work     k  = ,  consumed  in  stopping  the  body  in  the  direction 

2 

M  V'1 
A  G,  which  must  be  equal  to  the  work     lc'  = stored  in  the 

body  when  arriving  at  A,  or 

FR    MV1  ,    MV* 

= .  of  which  the  centripetal  force     F= , 

22  R 

and  which  is  the  solution  of  the  problem. 

The  forces  b,  d  and  f,  acting  in  the  direction  parallel  to  A  IT,  have 
given  the  body  the  velocity  I7 when  arriving  at  JE,  or  if  that  variable 
force  had  been  applied  on  the  body  at  rest  at  A  in  the  direction  A  H, 
it  would  have  produced  the  velocity  V  at  H.  Set  off  the  forces 
F,  b,  d  and/  at  right  angles  to  the  line  A  JT,  and  at  the  respective 
positions  of  the  body.  Join  these  forces  with  n  and  IT,  which  will  be  a 
straight  line.  The  area  of  the  triangle  An  H  represents  the 

Work     Jfc- 


accomplished  by  the  variable  force  acting  in  the  direction  parallel  to 
A  H,  and  which  is  equal  to  the 

_,    .      „     MV* 
Work     #  = • 

stored  in  the  body  when  arriving  at  E  or  H. 


CENTRIFUGAL  FORCE. 


225 


The  centripetal  force  is  equal  to  the  centrifugal  force  of  the  revolv- 
ing body,  or  the  force  of  inertia  presented  to  the  change  of  direction 

of  motion,  is  equal  to  the  centrifugal  force     F 


. 
R 

It  would  appear  from  this  formula  that  the  centrifugal  force  is  not 
a  simple  element,  but  it  will  be  evident  by  the  illustration  that  the 
spaces  A  G  or  A  If,  which  is  the  radius  of  the  circle,  is  accomplished 
by  the  product  of  velocity  and  time,  or  R  =  v  T,  in  which  v  is  the 
mean  velocity  in  the  time  T  or  space  R,  and  the  work  M  V1  is  di- 
vided by  the  space  JK,  which  gives  the  simple  element  force.  When 
the  mass  is  expressed  by  weight  the  centrifugal  force  will  be 


gR 

The  centrifugal  force  F  is  to  the  weight  W  of  the  revolving  mass, 
as  double  the  height  of  fall  due  to  the  velocity  V  is  to  the  radius  R 
of  revolution.  Call  8=  space  of  fall  due  to  V,  then 


T72 
=  —  :R. 


Velocity 

n    .  ., 
Centnfugal  force 

Weight  of  body 
Radius  of  revolution 

Velocity 


,     of  which    8-—. 


WV* 
jP=  JLJ-  ^  before  proved. 


=  0.10472  R 


2  TT  R 


,     2933.5  F 
R  =  —  - 


3 
4 
5 
o 


V293 
— 


WF? 


=  time  in  seconds  of  one  revolution. 

1.2272  W  R 


2933.5  T2 


226  ELEMENTS  OF  MECHANICS. 

Example  1.  A  body  weighing  W=  160  pounds  is  revolving  at  the 
rate  of  n  =  120  revolutions  per  minute  on  a  radius  R  =  1.5  feet  from 
centre  of  rotation  to  centre  of  gravity  of  the  body  (not  centre  or 
radius  of  gyration).  Required  the  centrifugal  force? 

,    160xl.5xl202 
F=        2933.5 

"  Example  2.  The  radius  of  the  earth  at  the  equator  is  about 
R  =  20,900,000  feet,  and  it  makes  one  revolution  in  24  hours  =  1440 
minutes  =  86400  seconds  =  T.  Required  the  centrifugal  force  of  a 
mass  at  the  equator  weighing  W=  2240  pounds  or  one  ton  ? 

,    1.2272x2240x20,900,000 

F= • — =  7.6963  pounds. 

86400s 

The  radius  of  the  earth  in  any  latitude  L  is 

^  =  20887680(1+0.00164  cos.2Z). 

The  centrifugal  force  at  any  point  on  the  surface  of  the  earth  will  be 
=  W(l  +  0.00164  cos.2L)cos.L 
2905.5 

The  vertical  action  of  this  centrifugal  force  will  be  as  the  cosine  of 
the  latitude. 

The  deviation  v  of  a  plumb-line  from  the  vertical,  caused  by  cen- 
trifugal force  of  the  earth's  rotation,  will  be 

Fsin.L 
ten.v-—^- 11 

The  deviation  of  the  surface  of  a  liquid  from  the  true  horizon  can 
be  calculated  from  the  same  forumlas. 

\  158.  CENTRIFUGAL  FORCE  OF  A  BODY  MOVING  ON  THE  SUR- 
FACE OF  THE  EARTH. 

A  body  moving  on  the  surface  of  the  earth  loses  in  weight  the 
faster  it  moves,  and  it  may  move  so  fast  that  it  loses  all  its  weight, 
which  happens  when  the  centrifugal  force  is  equal  to  the  weight  or 
force  of  attraction. 

TFV 
Centrifugal  force,          F= ,  in  which 

ff  -R 

W  •=  weight  of  the  moving  body  in  pounds. 
v  =  velocity  of  the  body  in  feet  per  second. 
R  =  radius  of  the  earth  in  feet. 


CENTRIFUGAL  FORCE.  227 

Example  3.  A  railway  train  weighing  1000  tons  or  W=  2240000 
pounds  is  running  on  a  horizontal  track  at  the  rate  of  60  miles  per 
hour,  or  velocity  v  =  88  feet  per  second. 

Required  the  centrifugal  force  of  the  train  in  the  curvature  of  the 
earth? 

„      Wv*       2240000  x88* 


The  train  of  1000  tons  lost  only  25  pounds  by  the  centrifugal 
force. 

This  centrifugal  force  is  without  regard  to  the  earth's  rotation 
around  its  axis. 

The  centrifugal  force  of  a  body  moving  on  the  surface  of  the  earth 
in  regard  to  the  earth's  radius  depends  upon  the  direction  of  motion. 
When  moving  from  east  to  west  there  is  less  centrifugal  than  when 
moving  from  west  to  east. 

The  centrifugal  force  of  a  body  at  rest  or  in  motion  on  the  earth's 
surface  affects  the  weight  of  that  body  when  weighed  on  a  spring 
balance. 

When  the  centrifugal  force  of  the  moving  body  is  equal  to  the 
weight,  or  F  =  W,  then  g  R  =  v2,  of  which  the  velocity  v  =  \/g  R. 

The  radius  of  the  earth  is  about  3956  miles  or  20887680  feet,  which 
inserted  in  the  formula  will  give  a  velocity 


v  =  -1/32.17  x  20887680  =  29528  feet,  or  5.5925  miles  per  second. 

That  is  to  say,  a  body  moving  with  a  velocity  of  5.6  miles  per 
second  at  or  near  the  surface  of  the  earth  would  not  fall  to  the  earth, 
because  its  centrifugal  force  is  equal  to  its  weight. 

A  cannon-ball  fired  horizontally  from  the  top  of  a  high  mountain, 
and  with  a  velocity  of  5.6  miles  per  second,  would  continue  to  rotate 
around  the  earth  and  make  one  revolution  in  about  seven  minutes 
and  a  half. 

This  is  the  principle  upon  which  the  moon  rotates  around  the 
earth — namely,  that  the  centrifugal  force  of  the  moon  is  equal  to  the 
centripetal  force  or  force  of  attraction  between  the  two  bodies. 

The  moon,  however,  does  not  revolve  around  the  earth's  centre,  but 
around  the  common  centre  of  gravity  of  the  two  bodies,  which  is  at 
about  0.7  of  the  earth's  radius  from  the  centre  of  the  earth. 


228  ELEMENTS  OF  MECHANICS. 


\  159.  CENTRIFUGAL  FORCE  OF  A  ROUND  DISK  ROTATING 
AROUND  ITS  CENTRE. 

The  dotted  line  represents  any  diameter  dividing  the  disk  into  two 
equal  parts.     When  rotating  each  part  tends  to  separate  in  that  di- 
„.  ameter  with  a  force  equal  to  the  centrifugal  force  of 

each  half.  The  radius  of  the  centrifugal  force  is 
the  distance  from  the  centre  of  rotation  to  the  cen- 
tre of  gravity  of  each  half.  The  centre  of  gravity 
of  a  semicircular  plane  is  0.424  of  the  radius  from 
the  centre. 

W=  weight  in  pounds  of  half  the  disk,  which 

may  be  of  any  thickness,  say  a  cylinder. 
It  =  outside  radius  in  feet  of  the  disk  or  cylinder. 
n  =  revolutions  per  minute. 


n    .  -A      i  . 

Centnfugal  force,        JP-___.     .         .     1 


-D      i  .-  2933.5  .F 


Example  1.  A  grinding-stone  of  6  feet  in  diameter  and  1  foot 
thick  is  making  n  =  100  revolutions  per  minute.  Required  the  cen- 
trifugal force  ? 

The  specific  gravity  of  sandstone  is  about  2.5,  when  the  weight  of 
the  grinding-stone  will  be 

62  x  0.785  x  62.33  x  2.5  =  4405.8  pounds.         F=  2202.9. 

Radius  of  grindstone  3  feet,  and  3x0.424  =  1.272  feet,  the  centrif- 
ugal radius. 

T    2202.9  x  1.272  x  100'     ft,,oc 

Centrifugal  force,         F  --     -  =  9o53.6  pounds. 
2933.5 

Example  2.  The  tensile  strength  of  ordinary  sandstone  may  be 
limited  to  100  pounds  per  square  inch  of  section.  Required  the  num- 
ber of  revolutions  per  minute  of  the  grinding-stone  in  the  preceding  ex- 
ample, at  which  the  centrifugal  force  would  break  it  ? 

The  section  in  the  diameter  of  the  stone  is  72x12  =  864  square 
inches,  which  multiplied  by  the  tensile  strength  100  will  be  F=  86400. 


^      .   .,  2933.5x86400     _m 

Revolution,     n  =  i  2?"  =         per  mmute" 


CENTRIFUGAL  FORCE. 


229 


§160.     CENTRIFUGAL   FORCE   OF   A    RING. 

The  centrifugal  of  a  ring  of  square  or  rectangular  section  is 

f  = 1 1  Fig.  186. 

2933.5 
TF=half  the  weight  of  the  ring. 


Example  1.  The  ring  of  a  cast-iron  fly-wheel  is  9  inches  square, 
the  outer  radius  It  =  6  feet,  and  r  =  5.25,  revolutions  per  minute 
n  =  64.  Required  the  centrifugal  force  of  the  fly-wheel  ? 

The  weight  of  the  ring  will  be 

(113  -  86.59)0.75  x  450  =  8910,         TF=  4455  pounds. 


Example  2.  The  tensile  strength  of  cast-iron  is  about  18000  pounds 
to  the  square  inch,  which  multiplied  by  81  square  inches,  the  section 
of  the  ring,  will  be  1458000  x  2  =  2916000  pounds,  the  force  F.  Re- 
quired the  number  of  revolutions  of  the  fly-wheel  in  the  preceding 
example  at  which  it  would  break  by  the  centrifugal  force  ? 


„      ,    . 
Revolution 


2933.5x2916000 


._n 

490  per  minute. 


§161.     THE   CENTRIFUGAL   GOVERNOR. 

The  action  of  a  governor  is  that  the  weight  of 
each  ball  is  to  the  centrifugal  force  as  the  height  h 
is  to  the  radius  of  rotation  r. 

W:F=h:r. 


Fig.  187. 


The  angle  x  which  the  arms  form  with  the  centre-shaft  is 
r     F  r 


230 


ELEMENTS  OF  MECHANICS. 


Centrifugal  force 
of  which 

p     >'  rn 

W^. 

x 

2933.5 

/  2933.5 

tan'X     51ieJto7i< 

\ 
54.16 

oa.io^ 
54.16 

O 

These  formulas  give  the  number  of  revolutions  per  minute  of  the 
governor  when  the  angle  x,  height  h,  radius  r  or  length  of  the  pen- 
dulum are  given. 

The  weight  of  the  balls  does  not  influence  the  angle  of  the  governor, 
because  the  centrifugal  force  varies  as  the  weight.  The  weights  only 
serve  to  do  the  work  of  regulating  the  steam-valve  when  the  velocity 
of  rotation  changes. 

Exampk.  The  pendulum  arms  of  a  governor  are  1=2  feet  from 
the  upper  joint  to  the  centre  of  the  balls.  How  many  revolutions  n  ? 
must  the  governor  make  per  minute  to  form  an  angle  x  =  45°  ? 

54.16 

4o  revolutions. 


|162.     THE   VARIABLE   PENDULUM   GOVERNOR. 

In  this  governor  the  balls  are  hung  at  a  distance  from  the  centre- 
line, by  which  the  length  of  the  pen- 
dulum varies  with  the  angle  x.  The 
distance  from  the  centre  a  of  the  ball 
to  where  the  direction  of  the  arm  cuts 
the  centre-line  at  b  is  the  real  length 
of  the  pendulum  in  that  position  of 
the  governor.  When  the  balls  hang 
vertical,  the  length  of  the  pendulum 
is  infinite. 

d  =  right-angular  distance  from  point 
of  suspension  to  centre-line  in 

Vfeet. 
e  =  distance  from  centre  of  ball  to 

point  of  suspension. 

Then  the  length  of  the  pendulum  in  any  position  of  the  balls  will 
be  in  feet,  l=d  cosec.x  -f  e. 

The  formulas  and  table  for  the  ordinary  centrifugal  governor  will 
also  answer  for  this  governor,  only  that  the  pendulum  length  must  be 
measured  from  a  to  b. 


GOVERNORS. 


231 


Revolutions  per  minute  of  governors  with  different  angles 

and  lengths  of  the  pendulum-arms. 

Length 

ANGLE  OF  PENDULUM  IN  DEGREES. 

I 

2O° 

25°  3O° 

35° 

4O° 

45° 

50° 

55° 

60° 

inches. 

n 

n 

n 

n 

n 

n 

n 

n 

n 

1 

193 

197 

202 

207 

214 

223 

234 

248 

265 

2 

137 

140 

143 

147 

152 

158 

166 

175 

187 

3 

112 

114 

116 

120 

124 

129 

135 

143 

153 

4 

97 

99 

102 

104 

108 

112 

117 

124 

133 

5 

87 

88 

90 

93 

96 

100 

105 

111 

119 

6 

79 

81 

83 

85 

88 

91 

96 

101 

109 

7 

73 

75 

76 

78 

81 

84 

89 

94 

100 

8 

68 

70 

71 

73 

76 

79 

83 

88 

94 

9 

64 

66 

67 

69 

71 

74 

78 

82 

88 

10 

61 

62 

64 

66 

68 

70 

74 

78 

84 

11 

58 

59 

61 

63 

65 

67 

70 

75 

80 

12 

56 

57 

58 

60 

62 

64 

67 

72 

77 

15 

50 

51 

52 

53 

55 

58 

60 

64 

69 

18 

45 

46 

47 

49 

51 

53 

55 

58 

63 

21 

42 

43 

44 

45 

47 

49 

51 

54 

58 

24 

39 

40 

41 

42 

44 

46 

48 

50 

54 

27 

37 

38 

39 

40 

41 

43 

45 

48 

51 

30 

35 

36 

37 

38 

39 

41 

43 

45 

48 

33 

34 

35 

36 

37 

38 

39 

41 

43 

46 

36 

32 

33 

34 

35 

36 

37 

39 

41 

44 

39 

30 

31 

32 

33 

34 

35 

37 

39 

41 

42 

30 

31 

32 

33 

34 

35 

36 

38 

41 

45 

29 

30 

31 

32 

33 

34 

35 

37 

40 

48 

28 

29 

30 

31 

32 

33 

34 

36 

39 

Vertical  height  h  in  inches  of  the  centre  of  suspension 

above  the  centre  of  the  balls,  of  governors  making  u 

revolutions 

per  minute. 

n 

h 

n 

h 

n 

h 

n 

h 

20 

88.00 

32 

34.38 

51 

13.53 

114 

2.71 

21 

79.82 

33 

32.32 

54 

12.07 

120 

2.44 

22 

72.73 

34 

30.45 

57 

10.83 

126 

2.22 

23 

66.54 

35 

28.74 

60 

9.78 

132 

2.02 

24 

61.07 

36 

27.16 

66 

8.08 

138 

1.85 

25 

56.32 

37 

25.71 

72 

6.79 

144 

1.69 

26 

52.07 

38 

24.38 

78 

5.78 

150 

1.56 

27 

48.29 

39 

23.14 

84 

4.54 

162 

1.34 

28 

44.90 

40 

22.00 

90 

4.34 

174 

1.16 

29 

41.86 

42 

18.18 

96 

3.82 

186 

1.01 

30 

39.11 

45 

17.38 

102 

3.38  i 

198 

0.88 

31 

36.63 

48 

15.27 

108 

3.02  i 

200 

0.87 

232 


ELEMENTS  OF  MECHANICS. 


Fig.  189. 


§  163.    ISOCHRONOUS  GOVERNOR  (Devised  by  the  Author). 

The  construction  of  this  isochronous  governor  is  readily  understood 
by  the  illustration.     It  is  perfectly  bal- 
anced, and  will  work  equally  well  in  any 
position  (horizontal,  inclined,  or  vertical) 
it  may  be  placed.     It  is  independent  of 
the  force  of  gravity. 
W=  weight  in  pounds  of  the  four  balls. 
F=  centrifugal  force. 
/=  force  in  pounds  on  the  spring  in  the 

direction  of  the  spindle. 
Z>  =  diameter  in  feet  of  the  circle   de- 
scribed by  the  centre  of  the  balls. 
d  —  length  of  the  two  levers,  as  shown 

on  the  drawing. 
e  =  distance  between  the   balls   in  the 

direction  of  the  spindle. 
The  centrifugal  force  of  the  four  balls 
will  be 

WDri* 


F- 


5867 


F:f-d:e, 


of  which    /= . 


WDerf      and 
/"   5867  d  ' 

The  force/  of  the  spring  ought  to  be  so  adjusted  that  the  balls  will 
form  a  square  when  the  governor  runs  at  the  average  speed. 

§  164.  CENTRIFUGAL  FORCE  OF  BODIES  MOVING  ON  CURVED 
ROADS. 

Fig.  190  represents  a  section  of  a  circus  ring,  and  a  rider  on  a  horse. 
It  is  well  known  that  the  faster  the  horse  runs  the  more  he  leans  to- 
ward the  centre  of  the  ring. 

Let  the  body  B  be  suspended  from  a  by  the  line  B  a,  and  swung 
around  the  circle  in  the  same  path  and  with  the  same  velocity  as  that 
of  the  horse  ;  then  the  inclination  of  the  line  B  a  will  be  the  same 
as  that  of  the  horse  and  the  rider.  From  the  formulas  for  the  centrif- 
ugal governor  we  have  the  angle  x  as  follows : 

Rn* 


MOMENTUM-CUR  VE8.  233 


R  =  radius  in  feet  of  the  centre  of  the  track. 

n  =  number  of  turns  around  the  circle  per  minute. 

The  rise  of  the  track  above  horizon  will  be  the  same  as  the  angle  x. 

Let  F  denote  the  velocity  in  feet  per  second  of  the  horse  or  ball. 

Kg.  190. 

a 

60 

n  =  ^0_F 
2*  R 

R  60'  F' 


tan.x  = 


2933.5(2  *  Ef 

3600  F2 


tan.x 


2933.5  x  4  x  9.86955  R     32.17  R 


This  formula  gives  the  angle  x  when  the  radius  R  and  velocity  F 
are  given.  When  we  have  the  velocity  expressed  in  statute  miles  per 

.  ,          ... ;  Miles2 

hour,  the  angle  x  will  be  :  tan.x  = 6 

14.956  R 

In  railway  curves  the  outer  rail  should  be  elevated  an  angle  x 
above  the  inner  rail.  Call  G  =  width  of  gauge  in  inches,  h  =  eleva- 
tion of  the  outer  rail  in  inches. 

h        Miles2                                  O  Miles2 
7  and.  o.       n  =  • -. 


G     14.956  R  15  R 

This  formula  is  not  strictly  correct,  because  —  is  sin.x,  instead  of 

G 

fan.z,  but  the  difference  is  of  no  practical  importance  in  the  small 
angle  of  elevation  of  the  outer  rail  in  railroad  curves. 

Example.  A  railroad  train  is  to  run  at  the-  rate  of  30  miles  per 
hour  on  a  curve  of  .#  =  500  feet  radius,  anjfc- 'gauge  (9  =  56.5  inches. 
Required  the  elevation  of  the  outer  rail  ? 

.     56.5x30' 

h  =  • • — •  =  b.o  inches. 

15  x  500 

With  this  elevation  the  wheels  would  bear  equally  on  both  rails. 
For  slower  speed  on  the  same  curve  the  wheels  would  bear  heavier 
on  the  inner  rail  than  on  the  outer  one. 


234 


ELEMENTS  OF  MECHANICS. 


Fig.  191. 


d 


\  165.     TO  MEASURE  ANGULAR  VELOCITY  BY  CENTRIFUGAL  FORCE. 

The  illustration,  Fig.  191,  represents  a  glass  tube  bent  into  the 
shape  of  a  fork  a,  b,  c,  d,  and  filled  with  mercury  to  the  line  ef. 

The  leg  c  d  is  set  into  rotation  around  the  other  leg  a  b  as  axis, 
and  the  centrifugal  force  of  the  mercury  in  the  part  b  c  will  raise  the 
mercury  in  the  leg  c  d  and  lower  it  in  a  b  to  a  difference  h  of  level 
in  the  two  legs. 

A  =  area  of  the  cross-section  of  the  tube  in  square 
inches. 

h  =  difference  of  height  of  the  mercury  in  inches. 

R  =  radius  of  revolution  of  the  leg  c  d  in  inches. 

n  =  number  of  revolutions  per  minute. 

W=  weight  in  pounds  of  the  column  of  mercury 
of  the  height  A. 

w  =  weight  of  the  mercury  in  the  part  b  c. 

The  weight  of  a  cubic  inch  of  mercury  at  the 
temperature  of  60°  Fahr.  is  0.941  of  a  pound. 
The  weights,    W=  0.941  A  h,     and  w  =  0.941  A  R. 

The  centrifugal  force  of  the  mercury  w  will  be 


F= 


E 


70405' 


which  must  be  equal  to  the  weight  W  which  acts  as  centrifugal  force. 
Insert  the  values  of  W  and  w  in  the  formulas,  and  we  have 

0.941  A  l?ns 


0.941 


70405 


241.44 


241.44 


70405' 

The  differential  height  A  of  the  mercury  is  independent  of  the  sec- 
tional area  A,  which  can  be  irregular. 

The  parts  a  b  and  c  d  of  the  tube  should  be  parallel,  but  the  part 
b  c  need  not  be  at  right  angles  to  the  legs,  nor  need  it  be  straight,  but 
can  be  made  of  any  curve  or  shape. 

Example  1.     The  radius  between  the  centre  lines  of  the  legs  is 
R  =  3  inches,  and  makes  n  =  184  revolutions  per  minute.     Required 
the  differential  height  A  ? 
3*  x  84 


-  =  4.327  inches,  of  which  one-half,  or  2.1635  inches, 


70405 
will  fall  in  the  leg  a  b  and  rise  the  other  half  in  the  leg  c  d. 


ROTATION  GAUGE. 


235 


Fig.  192. 


Example  2.  How  many  revolutions  must  the  instrument  make 
with  a  radius  J?  =  5  inches,  to  raise  a  differential  column  of  h  =  16 
inches  ? 

241 44     

n  =  —    — 1/16  =  193.152  revolutions  per  minute, 
o 

When  the  areas  of  the  cross-sections  of  the  bore  in  the  two  legs  are 
alike,  the  column  of  mercury  will  sink  in  the  centre  leg  a  b  as  much 
as  it  rises  in  the  leg  c  d,  or  e  e  =//',  and  a  graduated  scale  could  be 
attached  to  the  centre  leg  to  indicate  the  number  of  revolutions  of 
the  instrument.  It  is  not  necessary  to  make  the  cross-sections  of  the 
legs  alike,  as  will  be  explained  in  the  following  section. 

|  166.     REVOLUTION    INDICATOR. 

This  instrument,  Fig.  192,  is  based  upon  the  principles  described 
in  the  preceding  section,  and  invented  and  patented  by  Edward 
Brown  of  Philadelphia. 

The  centre  tube  a  b  is  made  of  glass  and  con- 
tains the  mercury  which  communicates  with  the 
iron  tube  c  d.  When  the  indicator  stands  still 
the  mercury  level  is  at/  e. 

The  iron  tube  is  fitted  with  a  vessel  at  d,  suf- 
ficiently large  to  contain  the  whole  column  e  d 
of  mercury,  so  that  when  the  indicator  revolves 
with  its  highest  speed,  the  mercury  will  fall 
from  e  to  e'  in  the  glass  tube,  and  rise  to  a  small 
height  in  the  vessel  d.  The  part  g  is  only  a 
balance  rod  of  solid  iron. 

The  vessel  d  is  turned  inward  at  the  top  to 
prevent  the  mercury  from  splashing  out. 

The  graduated  scale  is  held  in  position  by  the 
framing,  as  shown  by  the  illustration. 

The  instrument  is  intended  for  indicating  the 

rate  revolutions  of  a  steam-engine  or  other  rotary  machine,  for  which 
purpose  it  is  not  necessary  that  the  indicator  should  make  the  same 
number  of  revolutions,  but  can  be  geared  to  run  with  any  desired 
speed,  and  the  scale  is  graduated  to  suit  the  gearing  and  indicate 
directly  the  number  of  revolutions  of  the  engine. 

A  =  area  of  cross-section  of  the  vessel  d,  and  a  =  that  of  glass  tube. 
h  =  difference  of  height  of  the  mercury. 

x  =  depth  to  which  the  mercury  sinks  in  the  glass  tube  in  inches. 
n  =  revolutions  per  minute  of  the  indicator. 

R  =  radius  of  rotation  of  the  iron  tube  c  d. 


236 


ELEMENTS  OF  MECHANICS. 


70405 


-— 1.        x  =  - 


0405/1  +  -\ 

Example.  The  areas  a  :  A  =  1  :  16,  and  R  =  4  inches,  making 
n  =  15  revolutions  per  minute.  Required  the  sinkage  x  of  mercury 
in  the  glass  tube  ? 


The  indicator  is  geared,  say  3  to  1  revolutions  of  the  engine  ;  the 
latter  will  make  50  turns  per  minute  when  the  mercury  sinks  4.8125 
inches  in  the  glass  tube.  The  divisions  on  the  scale  will  be  a  ge- 
ometrical progression,  or  as  the  square  of  the  revolutions. 

The  scale  can  thus  be  graduated,  but  number  the  divisions  so  as  to 
correspond  with  the  revolutions  of  the  engine. 


|167. 


CENTRIFUGAL  FORCE    OF    A   LIQUID    ENCLOSED    IN    A 
ROTATING   VESSEL. 


Fig.  193. 


Fig.  193  represents  a  cylindrical  vessel  of  radius  R  and  height  _£T, 
about  half  filled  with  any  liquid,  say  to  the 
line  ef.  When  the  system  is  set  into  rota- 
tion the  centrifugal  force  will  form  the  liquid 
into  a  complement  paraboloid,  or  the  section 
of  the  liquid  will  be  a  parabola,  in  accord- 
ance with  the  following  formula : 


h 


241.44 


241.44 


70405  n  r 

h  =  abscissa    and     r  =  ordinate. 
The  formulas  give  the  form  of  the   para- 
bola in  the  rotating  vessel. 

The  ribs  b  b  are  fastened  inside  to  make 
the  liquid  rotate  with  the  vessel. 

The  angle  z  at  any  point  of  the  parabola 

2  A 

with  the  ordinate  radius  r  will  be  tan.z  =  —  . 

r 

The  same  angle  will  be  formed  by  the  surface  of  the  mercury  in 
the  vessel  d,  Fig.  192. 

The  cubic  content  of  a  paraboloid  is  one-half  of  that  of  a  cylinder 
of  the  same  base  and  height ;  the  liquid  in  a  rotating  cylindrical  ves- 
sel (Fig.  193)  will  therefore  sink  in  the  centre  as  much  as  it  rides  on 
the  sides  until  the  inverted  vertex  of  the  parabola  reaches  the  bottom 
of  the  vessel. 


PENDULUM.  237 


§168.    PENDULUM. 

A  body  freely  suspended  above  its  centre  of  gravity  and  made  to 
swing  is  called  a  pendulum. 

Simple  Pendulum  is  the  combination  of  a  small  body  hung  on 
a  light  line  and  made  to  swing. 

Compound  Pendulum  is  a  rigid  system  of  bodies  suspended 
above  the  centre  of  gravity  and  free  to  swing. 

A  body  suspended  at  c  by  the  line  I  will  hang  perpendicular  under 
the  point  of  suspension,  and  the  combination  is 
called  a.  plumb  line.  Draw  the  body  aside  from 
b  to  a,  and  leave  it  free  to  the  action  of  gravity, 
which  will  draw  the  body  back  to  b.  The  prim- 
itive work  consumed  in  drawing  the  body  aside 
is  stored  into  it  by  the  force  of  gravity  which 
moves  the  body  from  a  to  b,  where  it  arrives 
with  a  velocity  equal  to  that  due  to  an  equal 
height  of  fall  S.  A  body  in  motion  cannot  be 
stopped  without  restoring  the  work  which  has 
set  it  in  motion,  for  which  resistance  is  required,  and  as  the  body  met 
with  no  resistance  at  b,  its  motion  will  be  continued  to  c.  Whilst 
moving  from  b  to  c  the  body  meets  the  resistance  of  gravity,  which 
has  discharged  the  work  when  arriving  at  c,  an  equal  height  8 
above  b.  As  the  body  is  free  to  move  in  the  arc  of  the  circle,  the 
force  of  gravity  will  draw  it  back  to  b,  and  the  motion  continue  to 
a,  where  again  it  will  be  drawn  back  to  b,  and  so  it  will  continue  to 
move  fore  and  back  for  ever  if  no  other  force  interferes  with  the  ope- 
ration. This  operation  is  called  oscillation  of  a  pendulum. 

U69.     DYNAMICS   OF   THE   PENDULUM. 

Let  a  body  M,  Fig.  195,  be  suspended  by  an  inflexible  rod  without 
weight  from  the  centre  C,  and  free  to  revolve  in  the  whole  circle 
around  that  centre. 

J?  =  radius  of  the  circle  in  feet. 

<f  =  angle  of  the  rod  with  the  vertical  at  any  position  of  the  body 
M  in  the  circle. 

The  body  is  so  supported  in  its  centre  of  gravity  that  when  it 
swings  it  will  not  rotate  with  the  rod ;  that  is  to  say,  the  axis  a,  b  will 
remain  vertical  in  any  position  of  the  body  in  the  circle.  Then  each 


238 


ELEMENTS  OF  MECHANICS. 


particle  of  matter  in  the  body  will  describe  equal  circles  of  radii  JR. 
Move  the  body  M  to  the  highest  position  on  the  circle  at  0,  where  it 
will  be  balanced  on  the  rod  which  takes  up  the  whole  weight  W,  so 
that  the  body  will  remain  stationary ;  but  the  slightest  force  tending 
to  move  the  body  toward  1  or  15  will  disturb  the  equilibrium,  and 
the  force  of  gravity  with  the  co-operation  of  the  rod  will  cause  it  to 
describe  a  circle.  Suppose  the  motion  to  be  from  0  toward  1,  and  let 
the  lines  W  represent  the  weight  of  the  body,  which  is  the  force  of 
gravity.  In  the  several  positions  1,  2,  3,  4,  etc.  of  the  body,  resolve 
the  weight  W  into  two  forces,  one  acting  in  the  direction  of  motion 
or  the  tangent,  and  one  parallel  with  the  radius  of  that  position.  The 
body  is  then  moved  only  by  the  force  ^acting  in  the  direction  of  the 
tangent,  and  the  other  force  acting  parallel  with  the  radius,  or  at  right 
angles  to  the  direction  of  motion,  has  no  effect  upon  the  body's  motion. 

F  =  Wsin.y. 


Draw  a  straight  line  A,  B,  Fig.  196,  equal  to  the  length  of  the 
circumference  of  the  circle,  and  divide  it  into  an  equal  •  number  of 
parts  as  the  division  of  the  circle.  At  each  division  set  off  the  corre- 
sponding force  F  at  right  angles  to  A,  B,  and  join  these  forces  with 
the  curve  B  b  D  c  A.  Now,  if  the  body  M  be  placed  at  B  and  the 
same  forces  F  acting  upon  it  in  the  direction  B,  A,  it  will  attain  the 
same  velocity  as  that  at  the  corresponding  divisions  in  the  circle. 

The  body  will  attain  its  greatest  velocity  at  Z>,  which  corresponds 
with  that  at  the  lowest  position  in  the  circle  ;  after  which  the  forces 
F  act  opposite  the  motion  and  stop  it  at  A,  which  corresponds  with 


PENDULUM.  239 


the  highest  position  0  in  the  circle.  The  work  accomplished  by  the 
forces  F  is  represented  by  the  area  of  the  figure  bounded  within  the 
curve  B  b  D  c  A  and  the  straight  line  A,  B. 

The  area  B  b  D  represents  the  work  of  drawing  the  body  from  0 
via  1,  2,  3,  etc.  to  8,  and  which  is  equal  to  the  weight  W  multiplied 
by  the  diameter  of  the  circle,  or 


in  which  V=  velocity  of  the  mass  M  when  passing  the  lowest  position 
at  the  division  8. 

This  work  is  stored  in  the  body,  and  cannot  be  taken  out  of  it  with- 
out realizing  an  equal  amount  of  work,  which  is  accomplished  by  the 
body  continuing  its  motion  in  the  circle  against  the  action  of  the  force 
of  gravity,  until  it  reaches  its  starting  position  at  the  highest  point  0. 

The  work  accomplished  by  the  force  of  gravity  at  any  position  of 
the  mass  M  is  the  weight  W  multiplied  by  the  vertical  space,  or 
sinus-versiis  of  the  angle  <p  moved  through  —  namely, 


K= 


The  velocity  with  which  the  body  passes  any  division  or  point  in 
the  circle  will  be 

W~E  sinv.? 


M 

W 
M=  — ,  which  inserted  in  Formula  3  will  be 

g 


When  the  body  has  moved  from  the  highest  to  the  lowest  position  it 
has  described  an  angle  <p  =  180°,  for  which  sinv.  =  2  and  the  velocity 


2170.     TIME   OF   OSCILLATION. 

When  the  body  passes  the  lowest  position,  say  between  the  7th 
and  9th  divisions,  the  force  curve  is  nearly  a  straight  line,  and  we 
can  therefore  assume  the  motive  force  to  be  in  proportion  to  the 
length  of  the  arc  from  the  lowest  or  8th  division  to  the  position  of  the 
body.  Then  when  the  body  is  oscillating  a  small  angle,  the  motive 
force  is  directly  as  the  space  from  the  lowest  position,  which  corre- 


240  ELEMENTS  OF  MECHANICS. 

spends  with  the  case  in  §  92,  in  which  it  is  proved  that  the  time  is 
independent  of  the  space  of  action,  or  that  the  body  will  oscillate 
equal  times  independent  of  the  angle  of  oscillation. 

From  Formula  3,  §  92,  we  have  the  time  of  half  an  oscillation  to  be 


For  a  full  single  oscillation,       T-2-J—  .....        2 

»  c/ 

It  is  assumed  in  this  formula  that  the  motive  force  is  a  function  of 
the  arc,  or  F-  C  8. 


of  wliich  the  constant     (7-  3 


, 
180  180  -n  R  <p 

The  force  F  for  one  oscillation  will  be 


Insert  this  value  of  F  in  Formula  3  —  namely, 
c=  180x4  ygM^g  M 

JT  R  <?  180  7T  7T*  E   ' 

Insert  this  value  of  the  constant  C  in  Formula  2,  and  the  time  of 
one  single  oscillation  will  be 


=  -A'— .  O 

4  g  M      \g 

The  length  JR  of  a  pendulum  oscillating  T  seconds  will  then  be 

E= V 7 

Example.     Kequired  the  length  of  a  pendulum  making  one  single 
oscillation  per  second  ? 


PENDULUM. 


241 


§171.     ELEMENTS   OF   THE    PENDULUM. 

L  =  length  in  inches  of  pendulum  from  centre  of  suspension  to  cen- 
tre of  oscillation.  In  the  simple  pendulum  the  centre  of 
oscillation  is  in  the  centre  of  gyration. 

n  =  number  of  oscillations  in  T  seconds. 

T=  time  in  seconds  in  which  the  pendulum  makes  n  single  oscil- 
lations. 


Simple   Pendulum. 

12  g  T'1     39.114  T2 
L  =  —  -  -  =  -  -  -  . 


Fig.  197. 


If  the  suspended  weight  is  spherical  of  radius  r  and    {  C 
I  =  length  between  centre  of  suspension  and  centre  of 
ball,  the  pendulum  length  L  will  be  L  =  j/F+fr*  in  inches. 

Delicate  pendulum  experiments  must  be  made  in  a  vacuum  in  order 
to  agree  with  the  formulas,  and  the  acceleratrix  g  must  be  calculated 
for  the  latitude  and  elevation  above  the  level  of  the  sea  where  the 
experiment  is  made. 

The  length  of  a  pendulum  in  feet  for  the  oscillation  of  seconds,  is 
equal  to  the  acceleratrix  g,  divided  by  TT*. 

The  following  table  gives  the  pendulum  length  in  inches,  oscillat- 
ing seconds  with  the  corresponding  value  of  the  acceleratrix  g  in 


varous     aces  : 


pl 


Locations. 

Pendulum, 

L. 

Generatrix, 
9- 

Latitude. 
D.  m.    a. 

\t  the  Equator 

39  0152 

32  0789 

0°   0'    0" 

\t  Washington  D  C  

39.0958 

32.1552 

38   53  23 

At  New  York 

39  1017 

32  1608 

40   42   40 

At  Mean  Radius  of  the  Earth... 

39.1270 
39  1398 

32.1808 
32  1912 

45   00   00 
51    31 

At  Stockholm 

39  1845 

32  2281 

59  21   30 

31  2147 

32  2528 

79 

242  ELEMENTS  OF  MECHANICS. 

The  length  of  the  pendulum  for  seconds,  at  the  level  of  the  sea  in 
any  latitude,  will  be 

Z  =  39.127 -0.09982  cos.2  latitude. 

When  the  latitude  is  over  45°  the  cosine  for  double  the  latitude 
will  be  negative,  which  multiplied  by  the  negative  coefficient  —  0.09982 
gives  a  positive  product  to  be  added  to  the  pendulum  length  39.127 
of  45°  latitude. 

§  172.     ELEMENTS  OF  THE  COMPOUND  PENDULUM. 

Fig.  198  represents  a  rod  suspended  at  o,  and  free  to  swing. 
Fi    198  z  =  distance  fr°m  the  centre  of  suspension  o  to  the  centre 

of  gravity  A.     As  the  rod  is  parallel,  its  centre  of 
gravity  A  will  be  in  the  middle. 

x  =  distance  from  centre  of  suspension  to  centre  of  gyra- 
tion J3,  which  is  the  radius  of  gyration. 
L  =  distance  from  the  centre  of  suspension  to  the  centre 
of  oscillation,  which  is  the  pendulum  length  of  the 
rod. 
These  distances  bear  the  following  relation  to  one  another : 

z  :  x  =  x  :  L. 

That  is  to  say,  the  radius  of  gyration  x  is  the  mean  pro- 
portion between  z  and  L.  This  proportion  will  hold  good 
for  any  kind  of  compound  pendulum. 

z  =  — ,       x  =  i/z  L,        L  =  — . 
L  z 

1  =  length  of  the  whole  rod. 

Centre  of  gravity     z  =  \  I. 

Radius  of  gyration  x  =  0.57735  I. 

Then  the  pendulum  length  of  a  rod  or  bar  will  be 

^,(057735^ 
z  0.5 1 

or  the  pendulum  length  is  f  of  the  whole  length  of  the  bar. 

In  a  compound  pendulum  the  particles  of  matter  located  near  to 
the  centre  of  suspension  have  a  tendency  to  oscillate  faster,  and 
those  at  the  greatest  distance  tend  to  oscillate  slower,  than  does  the 
compound  pendulum;  and  as  all  the  particles  are  rigid  into  one  body, 
there  must  be  one  of  them  which  has  no  tendency  to  oscillate  faster 
or  slower.  This  particle  is  called  the  centre  of  oscillation,  and  if  it 


PENDULUM. 


243 


was  suspended  alone  as  a  single  pendulum,  it  would  oscillate  the  same 
time  as  does  the  compound  pendulum. 

Professor  Huyghens  of  Holland  discovered  that  the  centres  of 
suspension  and  oscillation  are  convertible  into  one  another ;  which  is 
to  say  that  if  the  compound  pendulum  be  suspended  in  its  centre  of 
oscillation,  the  former  centre  of  suspension  will  then  be  the  centre  of 
oscillation,  and  it  will  oscillate  the  same  lengths  of  time. 

The  pendulum  length  of  a  compound  pendulum  can  thus  be  ascer- 
tained with  great  precision  by  suspending  it  in  two  different  points 
alternately,  so  that  it  will  oscillate  equal  lengths  of  time  in  both  cases. 

The  experiments  of  pendulum  oscillation  should  be  made  under 
vacuum,  in  order  to  avoid  the  influence  of  the  resistance  of  air. 

The  value  of  the  acceleratrix  g  can  be  determined  with  great  pre- 
cision by  a  pendulum  whose  length  is  accurately  known,  for  which 
purpose  a  compound  pendulum  in  form  of  a  parallel  rod  or  bar  ap- 
pears to  be  the  best. 

^  L  -*  n* 
9-  12  T,- 

t  173.     RADIUS  OF  GYRATION. 

The  radius  of  gyration  of  an  oscillating  body  can  be  accurately  de- 
termined by  the  number  of  oscillations  n  in  the  time  T. 

12 


6.2541 


-o  j-  .-  l^gT^z 

Eadms  gyration     x  =  \  —  -  = 
\     -2  n* 


$  174.  TO  FIND  THE  PENDULUM  LENGTH  OF  A  CYLINDER  SUSPENDED 
AT  ONE  END. 

Fig.  199. 
O 

L  =  length,    and    r  =  radius  of  cylinder. 
Centre  of  gravity        z  =  0.5  /. 

Radius  of  gyration     x  = 
Pendulum  L  = 


12 


244 


ELEMENTS  OF  MECHANICS. 


§175.     PENDULUM    LENGTH   OF   A   SUSPENDED   BALL. 

Fig.  200.  z  =  distance  from  centre  of  suspension  to  centre   of 

gravity  of  the  ball. 
r  —  radius  of  the  ball. 

Radius  of  gyration     x  =  j/z'  +  fr1. 

x2      z*  +  0.4  r1 
Pendulum  L  =  —  = . 


\  176.     COMPOUND   PENDULUM  OF  TWO  RIGID   BALLS. 

R  and  r  =  radii  of  the  balls. 

a  and  b  =  distances  from  centre  of  suspension  to  cen- 
tres of  gravity  of  the  balls. 

Pa+Qb 
Centre  of  gravity  z  = ^L—. 


P+Q 


+()A 


P(a'+() 


P+Q 


T 
Pendulum     L  =  —  = 

z 


Pa+Qb 


§177.     DOUBLE   COMPOUND   PENDULUM. 

Notation  of  letters  are  the  same  as  in  the  preceding 
paragraphs. 

Pa-Qb 

Centre  of  gravity          z  =     p+^    . 

Radii   of    gyration, 


r  enau  lum     Li  =  —  «= 

z 


Pa-Qb 


When  the  pendulum  length  is  given,  the  oscillations 
are  calculated  by  the  formulas  for  the  simple  pendulum. 


PENDULUM. 


245 


Example.     Weight  and  dimensions  of  the  double  compound  pendu- 
lum balls  are  as  follows  : 

Weight  P=50  pounds,  radius  JS  =  4  inches,  lever  a  =  25  inches. 
Weight  Q  =  20  pounds,  radius   r  =  3  inches,  lever  b  =  15  inches. 

Required  the  pendulum  length  Jj?   and   single   oscillations   per 
minute  n  ? 

T     50(25'+  0.4  x42)  +  20(15'  +  0.4  x32) 
Z  =  -  50x25-20xT5-       J 

Oscillations  per  minute,  §  171,  will  be 


V/37.845 

It  is  nearly  a  second  pendulum. 

The  nearer  the  double  compound  pendulum  is  suspended  from  its 
centre  of  gravity,  the  longer  will  be  the  pendulum  length. 

Pendulum  Lengths  for  Different  Numbers  of  Vibrations 
per  Minute,  and  Time  of  each  Vibration  in  Seconds. 


No. 

Feet. 

Time. 

No. 

Inches. 

Time. 

No. 

Inches. 

Time. 

1 

11734 

60 

28 

179.60 

2.15 

80 

22.001 

0.750 

2 

2935.5 

30 

29 

167.43 

2.08 

84 

18.181 

0.712 

3 

1303.7 

20 

30 

156.45 

2.000 

90 

17.384 

0.666 

4 

733.37 

15 

32 

137.51 

1.88 

96 

15.267 

0.625 

5 

462.36 

12 

34 

121.84 

1.77 

102 

13.534 

0.586 

6 

325.85 

10 

36 

108.65 

1.67 

108 

12.072 

0.555 

7 

239.47 

8.8 

38 

97.512 

1.58 

114 

10.835 

0.525 

8 

183.35 

7.50 

40 

88.005 

1.50 

120 

9.7785 

0.500 

9 

144.86 

6.66 

42 

79.825 

1.43 

132 

8.0815 

0.452 

10 

117.34 

6.00 

44 

72.731 

1.37 

144 

6.7900 

0.416 

11 

96.975 

5.45 

46 

66.545 

1.31 

156 

5.7860 

0.383 

12 

81.462 

5.00 

48 

61.071 

1.25 

168 

4.5452 

0.356 

13 

69.433 

4.62 

50 

56.325 

1.20 

180 

4.3460 

0.333 

14 

59.867 

4.30 

52 

52.075 

1.16 

192 

3.8168 

0.313 

15 

52.15 

4.00 

54 

48.290 

1.12 

204 

3.3835 

0.295 

16 

45.838 

3.76 

56 

44.900 

1.08 

216 

3.0180 

0.279 

17 

40.602 

3.53 

58 

41.860 

1.04 

228 

2.7087 

0.265 

18 

36.215 

3.34 

60 

39.114 

1.000 

240 

2.4445 

0.250 

19 

32.504 

3.17 

62 

36.631 

0.965 

252 

2.2173 

0.237 

20 

29.335 

3.00 

64 

34.379 

•  0.934 

264 

2.0204 

0.228 

21 

26.608 

2.86 

66 

32.326 

0.907 

276 

1.8485 

0.218 

22 

24.244 

2.73 

68 

30.450 

0.882 

288 

1.6975 

0.209 

23 

22.183 

2.62 

70 

28.737 

0.855 

300 

1.5646 

0.200 

24 

20.365 

2.50 

72 

27.162 

0.833 

324 

1.3444 

0.185 

25 

18.775 

2.41 

74 

25.714 

0.818 

348 

1.1627 

0.173 

26 

17.368 

2.32 

76 

24.378 

0.790 

372 

1.0175 

0.162 

27 

16.097 

2.23 

78 

23.144 

0.769 

396 

0.88105 

0.152 

21* 


246 


ELEMENTS  OF  MECHANICS. 


§178.     CONICAL   PENDULUM. 

The  swing  of  a  pendulum  can  be  made  to  describe  a  cone,  the  ver- 
tex of  which  is  in  the  centre  of  suspension,  and  the  base,  which  may 
be  a  circle  or  an  ellipse,  is  at  the  lowest  point  of  the  pendulum.  This 
is  called  a  conical  pendulum,  and  is  illustrated  by  Fig.  203. 

Let  the  body  b  be  suspended  from  o  and  made  to  revolve  in  the 
circle  a,  b,  c,  d;  then  the  height  ZTmay  represent  the  weight  W  of  the 
body,  radius  R  the  centrifugal  force  F,  and  L  the  strain  /  in  the 
direction  of  the  pendulum. 

x  =  angle  of  the  pendulum  with  the  axis  H. 


Then, 
WR 


- 


F:W=R:  H. 

¥  * 


WRn* 


2933.5 


WR 
H 


— —  =  —  of  which  n1  = 


Revolutions  per  minute,        n  =  -*/ 


WR 


2933.5 

H 
2933^ 

H 


This  formula  proves  that  the  number  of  revolutions  per  minute  are 
constant  for  any  constant  height  H,  and  independent  of  the  angle  x, 
which  will  be  understood  by  Fig.  204. 

The  circles  aa,bb  and  c  c  described  by  pendulums  L,  L',  L"  are 
in  the  same  plane,  and  H  the  height 
of  the  centre  of  suspension  o  above 
that  plane.  Any  length  of  pendulum 
L,  L' ,  L"  describing  a  circle,  a  cone 
of  any  constant  height  H  will  make 
the  same  periodical  revolutions  or  angu- 
lar velocity. 

Let  t  denote  the  time  in  seconds  of 
one  revolution  of  the  conical  pendulum. 


Then,    «-«?, 
n 


,         60 
and  n  =  — , 
t 


The  time  of  one  revolution  of  a  conical  pendulum  is  directly  as  the 
square  root  of  the  height  H. 


OXFORD  PENDULUM. 


247 


The  time  of  a  double  oscillation  of  a  pendulum  in  a  plane  is 


Fig.  205. 


The  time  of  one  double  oscillation  of  a  pendulum  in  a  plane  is 
directly  as  the  square  root  of  the  length  L. 

2179.     THE   OXFORD   PENDULUM. 

Fig.  205  represents  a  perspective  view  of  a  combination  of  two 
pendulums  of  different  lengths  invented  at  the  Oxford  University, 
England. 

A  line  or  wirey,  o,  g  is  fastened  at  f  and  g,  at  which  point  the  line 
is  free  to  swing.  Another  pendulum  o  e  is  hung  at  o,  so  that  of=  o  g. 
The  points  of  suspension  /  and  o 
should  be  in  a  horizontal  line 
f,  o' ,  g.  A  heavy  body  is  hung 
on  the  pendulum  at  e  to  form  the 
centre  of  oscillation.  A  rectan- 
gular screen  a,  c,  b,  d  is  placed 
with  its  centre  under  the  pendu- 
lum, so  that  the  centre  line  a  b  is 
parallel  with  f  g.  The  body  e  is 
a  funnel  filled  with  dry  black 
sand,  to  fall  on  the  screen  and 
mark  the  course  of  the  pendulum 
when  set  into  vibration. 

Place  the  funnel  at  a  and  let  it 
vibrate  the  angle  a,  o,  b,  when  the 
sand  will  mark  the  straight  line 
a  b  across  the  screen.  Place  the 
funnel  at  c  and  let  it  vibrate  the 
angle  c,  o' ,  d,  when  the  sand  will 
mark  the  straight  line  c  d. 

As  the  pendulum  o'  e  is  longer 

than  o  e,  it  requires  more  time  to  vibrate  from  c  to  d  than  it  does 
from  a  to  b. 

Now  place  the  funnel  at  the  corner  h  of  the  screen,  and  lea've  it  to 
its  own  course  of  vibration.  The  short  pendulum  o  e  will  vibrate 
across  the  screen  in  periods  of  time  equal  to  that  when  vibrating  from 
a  to  b,  whilst  the  long  pendulum  o'  e  will  vibrate  the  same  periods  as 
from  c  to  d;  the  combined  vibrations  of  the  two  pendulums  will  cause 
the  funnel  to  describe  a  regular  system  of  curves,  which  will  be  marked 
by  the  sand  on  the  screen.  The  form  of  the  figure  so  marked  de- 
pends upon  the  proportion  of  lengths  of  the  two  pendulums. 


248 


ELEMENTS  OF  MECHANICS. 


Fig.  207. 


Fig.  208. 


Figure  206.     L  :  1  =  4  :  1,     T:t  =  2:l. 

The  funnel  is  started  from  ra  with  its 
natural  course  of  vibration,  and  will  trace 
the  parabola  m  b  n.  On  arriving  at  n  the 
funnel  will  return  in  the  same  pata  via  b 
to  m,  and  repeat  continually  the  same  course 
shown  by  the  dashed  line. 

If  instead  of  allowing  the  funnel  to  take 
its  own  course  from  m,  it  is  started  by  a  lat- 
eral force  in  the  direction  of  the  dotted  line, 
the  funnel  will  describe  that  dotted  line  per- 
petually. 

The  funnel,  started  with  proper  velocity  in 
any  direction  of  the  black  line,  would  de- 
scribe that  figure  8  continually. 

A  variety  of  figures  can  thus  be  drawn 
with  a  constant  proportion  of  pendulums, 
only  by  starting  in  different  directions. 

Figure  207.     Z:/=16:9,     T:  i!  =  4  :  3. 

The  funnel  started  in  its  own  course  from 
m  will  describe  the  dotted  line,  and  finally 
arrive  at  n,  where  it  will  return  in  the  same 
path  to  m. 

The  dotted  line  is  the  figure  drawn  on  the 
screen  Fig.  205. 

If  the  funnel  is  started  with  proper  ve- 
locity in  any  direction  of  the  full-drawn  line 
it  will  describe  that  line  perpetually. 

It  is  almost  impossible,  or  rather  a  chance, 
to  start  the  vibration  with  such  precision  as 
to  trace  only  that  black  line  ;  but  in  practice 
the  funnel  will  describe  a  more  complicated 
and  better-looking  figure,  which  could  not  be 
constructed  without  extraordinary  labor  and 
patience. 

The  short  pendulum  will  make  4  oscilla- 
tions while  the  long  one  makes  3. 

Figure  2O8.     L  :  Z  =  25  :  9,      T:  t  =  5  :  3. 
This  figure  is  described  in  the  same  way 


PENDULUM  DIAGRAMS. 


249 


as  the  preceding  one,  only  with  different  pro- 
portion of  pendulums.  The  long  pendulum 
will  make  3  oscillations  while  the  shorter  one 
makes  5. 

Figure  209. 


This  figure  is  drawn  only  by  the  natural 
course  of  the  pendulum  starting  from  m  and 
ending  at  n. 

The  short  pendulum  will  make  11  oscilla- 
tions whilst  the  long  one  makes  7. 

Figure  21O. 

.£  :  £  =  529  :  169,         jP:  2-23: 13. 

The  funnel  is  left  to  take  its  own  course 
from  the  corner  m,  and  traces  the  figure  as 
shown  by  the  illustration.  When  the  funnel 
comes  nearest  to  the  corner  n' ,  it  will  return 
in  the  direction  of  the  dotted  line  and  trace 
another  figure  between  the  lines,  \intil  finally  UU 
it  arrives  at  the  corner  n,  when  the  whole 
figure  is  complete. 

The  short  pendulum  will  make  23  oscilla- 
tions whilst  the  long  one  makes  13. 

Figure  211. 

L  :  £  =  529  :  361,         T:  t  =  23  :  19. 

This  figure  is  similar  to  the  preceding  one, 
only  that  the  short  pendulum  makes  23  os- 
cillations whilst  the  long  one  makes  19. 

It  is  supposed  in  the  above  figures  that  the 
angles  of  vibration  of  the  two  pendulums 
are  constant,  which  is,  however,  not  the  case 
in  practice.  The  resistance  of  the  air  to  the 
funnel  gradually  diminishes  the  angles  of  vi- 
bration, so  that  the  funnel  can  never  return 
in  its  former  course ;  which  circumstance 
makes  the  figure  more  complicated  and  of 
a  curious  shading. 


71 


Fig.  211. 


250  ELEMENTS  OF  MECHANICS. 

It  is  very  difficult  to  represent  the  true  appearance  of  the  various 
figures  by  drawings. 

When  the  angles  of  vibration  are  much  reduced,  the  figure  becomes 
darkest  near  the  centre  of  the  screen,  and  in  the  crossings  the  sand 
is  piled  up  into  regularly-formed  heaps,  which  make  a  beautiful  ap- 
pearance. 

An  infinite  variety  of  figures  can  thus  be  made  by  tracing  one  on 
the  top  of  the  other. 

A  machine  could  easily  be  constructed  to  make  these  figures  per- 
fectly, and  which  could  be  so  arranged  that  a  particular  shading  of  a 
figure  could  not  be  reproduced  even  by  the  same  machine. 


§180.     CENTRE   OF   PERCUSSION. 

Take  a  bar  of  hard  materials  in  your  hands  and  strike  it  over  a 
sharp  edge,  then  if  no  shock  is  felt  in  the  hands,  the  bar  struck  with 
its  centre  of  percussion  over  the  edge.  If  a  downward  shock  is  felt, 
the  bar  struck  with  its  centre  of  percussion  inside  the  edge,  and  with 
an  upward  shock  outside  the  same.  The  work  stored  in  the  mov- 
ing bar  is  discharged  over  the  edge  and  in  the  hands.  If  the  mo- 
mentum in  the  moving  bar  is  not  evenly  divided  on  both  sides  of 
the  edge,  the  difference  will  be  felt  in  the  hands ;  but  if  evenly  divided, 
all  the  work  will  be  discharged  on  the  edge  and  no  shock  felt  in  the 
hands. 

Assume  two  bodies  M  and  m  hung  on  an  inflexible  line  without 
Fig.  212.  weight  and  suspended  at  the  point  C,  so  as  to 

form  a  pendulum.  The  edge  E  is  placed  at 
the  centre  of  percussion  of  the  bodies  M  and 
m,  so  that  when  the  pendulum  swings  and 
strikes  E,  there  will  be  no  shock  felt  in  the  cen- 
tre of  suspension  O.  The  condition  under 
which  this  can  be  accomplished  is  that  the 
momentums  of  M  and  m  must  be  equally  di- 
vided on  both  sides  of  E. 

v=  velocity  of  M,  and  v  that  of  m,  when  the 
edge  is  struck.     The  momentums  in  the  bodies 
will  then  be  M  V  and  m  v. 
The  letters  X,  I,  x,  b  and  c  are  as  represented  in  Fig.  209. 
When  the  bodies  are  moving  around  the  common  centre  C,  the 
velocities  Fand  v  will  be  as  their  radii  Xand  x. 


ORDNANCE  DYNAMICS.  251 

The  momentums  of  motion  will  then  be  M  X  and  m  x,  which  mul- 
tiplied by  their  respective  levers  of  action  b  and  c  must  be  alike. 

M  Xb  =  m  x  c. 
b  =  X-l,  and    c-l-z. 

Then  M  X(X-l)=m  x(l-x). 

MX^-MXl^mxl-mx*. 


1  = 


M  X+mx 


The  masses  M  and  m  can  in  this  formula  be  expressed  either  in 
masses,  weights  or  volumes. 

Example.  The  masses  M=  5  and  ra  =  4  pounds,  the  radii  of  gyra- 
tion X=  3  and  x  =  2  feet.  Required  the  length  I  from  the  centre  of 
oscillation  to  the  centre  of  percussion? 


5x3  +4x2 


= 


This  length  is  the  same  as  the  pendulum  length,  or  the  centre  of 
percussion  is  the  same  point  as  the  centre  of  oscillation. 

The  centre  of  percussion  is  therefore  calculated  by  the  same  for- 
mulas as  for  the  pendulum  or  centre  of  oscillation. 

A  body  or  a  system  of  bodies  suspended  in  its  centre  of  percussion 
will  have  its  new  centre  of  percussion  in  the  former  centre  of  sus- 
pension. 

§181.    ORDNANCE    DYNAMICS. 

Let  a  spring  P  of  force  F  be  ap- 
plied  between  two  masses  M  and  ra, 
free  to  move.  As  the  forces  F  act- 
ing in  opposite  directions  are  alike, 
the  spaces  R  and  8  and  the  velocities 
V  and  v  will  be  inversely  as  the 
masses. 

R:8=M'.m,      .         .1 

V:v=M:m,      .         .2 

and       R  :  8=  V  :  v.  .3 


252 


ELEMENTS  OF  MECHANICS. 


If  the  force  F  is  constant  through  the  spaces  R  and  S,  the  veloci- 
ties will  be 

ITF3F 
IF 4 


The  work  stored  in  each  mass  will  be 

K-fS-Xll. 


7  =  F  o  =  m  v*  7 

o          ^    . 

In  the  case  of  firing  a  gun,  M  represents  the  mass  of  the  gun  and 
gun-carriage,  m  the  mass  of  the  ball  and  F  the  force  of  the  gun- 
powder. The  length  of  the  bore  of  the  gun  passed  through  by  the 
ball  is  I =£+£>,  and  R  is  the  recoil 
of  the  gun. 

The  recoil  is  generally  partly  coun- 
teracted by  a  force/  applied  on  MAS 
friction  against  the  force  F,  as  repre- 
sented by  Fig.  210.  . 


Fig.  214. 


(F-f)  :  M  =  V:  T 

F:m  =  v  :  T 

F  MV 


MV      mv 
\F-f)~~F' 


m(F-f)1 


~Y= 


M  F 

MV 

ofwhlch    r  = 


of  which    S= 


2  £ 


mv2      MV* 

•  = +. 


2(F-f) 


.     8 
.     9 

.  10 

.  11 
.  12 


Recoil 


F  M 


-+1 


.  14 


ORDNANCE  DYNAMICS. 


253 


This  is  the  recoil  at  the  moment  the  mass  or  ball  m  reaches  the 
space  S,  but  the  mass  J/has  then  a  velocity  V,  which  must  be  stopped 
by  the  force  f  in  an  additional  recoil  r' . 

I  MV* 

-,  and  the  whole  recoil   r+i/  =  —    ^ +     .   .  .     15 


F  M 


This  formula  reduces  itself  to 

[r,          IFm(F-f) 


.  16 


It  is  supposed  in  the  above  arguments  that  the  mass  or  'ball  m 
moves  through  the  bore  without  friction  or  other  resistance,  which 
cannot  be  the  case ;  and  when  that  friction  is  taken  into  account  the 
recoil  will  be  diminished  considerably.  This  friction  and  resistance 
to  the  ball  in  the  bore  acts  to  drag  the  gun  with  it,  so  that  there 
may  be  ho  recoil  until  the  ball  leaves  the  muzzle,  as  has  been  con- 
firmed by  experiments  with  the  ballistic  pendulum. 

'i  182.     THE  BALLISTIC  PENDULUM. 

This  pendulum,  represented  by  Fig.  215,  is  designed  for  the  pur- 
pose of  measuring  the  velocity  and  work 
of  a  body  striking  it.  It  consists  of  a 
long  rod  suspended  at  one  end,  and  with 
a  block  of  wood  or  some  other  soft  ma- 
terial at  the  other  end,  forming  a  pen- 
dulum. 

The  point  b  is  the  centre  of  gravity  of 
the  whole  pendulum,  and  c  the  centre  of 
oscillation. 

A  body,  or  say  a  rifle-ball  J3,  striking 
the  wood  block  will  move  the  pendulum 
an  angle  x  and  raise  the  centre  of  gravity 

a  space  s,  which  is  the  versed  sine  of  x.  The  body  B  should  strike 
the  pendulum  in  a  horizontal  direction  toward  the  centre  of  oscilla- 
tion c,  in  order  to  avoid  jarring  in  the  fulcrum  a. 

The  centre  of  oscillation  should  be  determined  by  allowing  the 
pendulum  to  swing  and  counting  the  number  of  single  oscillations  n 
per  time  t  in  seconds.  The  pendulum  length  Z/,  from  the  fulcrum  a 
to  the  centre  of  oscillation  c,  will  be 

.     39.114  f*  . 

2j  = ,  in  inches. 

n2 


Fig.  215. 

_, 

1 

V 

I 

IS 

Aas. 

i  \ 

'—  j—" 

!*li.-    \ 

254  ELEMENTS  OF  MECHANICS. 

The  centre  of  gravity  b  is  found  by  balancing  the  pendulum  over  a 
sharp  edge  or  by  suspending  it  in  different  positions,  as  described  on 

50. 

The  work  of  raising  the  pendulum  the  vertical  space  s  is  equal  to 
the  work  of  the  striking  body  B. 

W=  weight  of  the  pendulum  in  pounds. 
s  =  space  in  feet  which  the  pendulum  is  lifted. 
B  =  weight  of  the  striking  body  in  pounds. 
V=  striking  velocity  in  feet  per  second. 

The  work          Ws  = 


*9 
The  velocity 

The  space  s  =  jR  ver.$in  x. 

The  angle  x  is  measured  by  a  graduated  arc  under  the  pendulum. 

Example.  The  weight  of  the  pendulum  is  W=400  pounds,  and 
the  distance  from  the  fulcrum  to  the  centre  of  gravity  JK  =  10  feet. 

The  weight  of  the  striking  body  is  _Z?  =  0.08  of  a  pound,  which 
moves  the  pendulum  an  angle  ar  =  36°  53'.  Required  the  striking 
velocity  of  the  body  B  ? 

Space  s  =  10  x  arer.sm.360  53'  =  2  feet. 


Velocity      F=  J2*32-17*400*2  =  §02.3  feet  per  second. 
»  0.08 


.  DYNAMICS  OF  HEAVY  ORDNANCE. 

The  force  of  ignited  gunpowder  enclosed  in  a  gun  varies  with  the 
quickness  of  the  powder,  and  has  been  found  to  reach  as  high  as  40 
tons  to  the  square  inch.  The  work  of  gunpowder  in  heavy  ordnance 
expressed  by  the  ordinary  unit  foot-pound  becomes  a  very  high  num- 
ber, for  which  a  larger  unit  has  been  adopted  by  English  ordnance 
officers  —  namely,  that  of  foot-ton,  which  means  a  work  of  lifting  one 
ton  of  2240  pounds  one  foot  high.  This  unit  reduces  considerably 
the  number  which  expresses  the  work  of  heavy  ordnance. 

Let  M  denote  the  mass,  and  TFthe  weight  in  pounds  of  a  projectile 
of  velocity  V  feet  per  second  ;  then  the  work  stored  in  the  projectile 
will  be,  as  before  proved, 


,    MV*  . 

K=*  —  -  —  =  -  in  foot-pounds. 


ORDNANCE  DYNAMICS.  255 

W  V1  WV* 

K  = = in  toot-tons.       .         .     2 

2^x2240     144121.6 

144121.6  k 
Weight  of  projectile  W=- — —pounds.    .     3 

Velocity  of  feet  per  second      V=  -*/ : .    .         .     4 

The  dynamic  work  of  different  kinds  of  gunpowder  utilized  in 
heavy  ordnance  varies  between  60  and  90  foot-tons  per  pound  of 
powder.  The  average  may  be  taken  to  be  80  foot-tons.  Let  P  denote 
the  weight  in  pounds  of  powder  in  a  charge,  then  the  work 


W  F2 
£  =  80P= 


Weight  of  charge 


Weight  of  projectile      TF= 


144121.6 
IFF2 


11,500,000 
11, 500,000  P 

v* 


TT,    -.      t       •    .-i      v       /11,500,OOOP  Q 

Velocity  oi  projectile     l/  =  \/ •        ,'•'•        •        •     " 

Example.  A  gun  is  loaded  with  a  charge  of  P  =  30  pounds  of 
powder  for  a  projectile  weighing  TF=200  pounds.  Required  the 
muzzle  velocity  of  the  projectile? 


„         .     .0      „       111,500,000x30     1Q1Q,    .  ,       . 

Formula  49.      r  "••%/  ---  --  =  1313  feet  per  second.     Ans. 

*  ^00 

Quick  powder  in  small  firearms  utilizes  less  work  per  weight  of  the 
explosive  than  does  slow  powder  in  heavy  ordnance.  The  formulas 
from  47  to  49  inclusive  are  equally  applicable  for  small  firearms,  in 
which  the  weights  of  the  powder  and  projectile  are  expressed  in 
grains. 

Coefficients,  j  11'500'000  for  heav?  ordnance' 
i    8,000,000  for  small  firearms. 


§184.     DYNAMIC    DIAGRAMS   OF   HEAVY   ORDNANCE. 

The  following  diagrams  are  deduced  from  English  experiments 
with  heavy  ordnance  made  in  the  year  1870.  The  experiments  were 
made  by  an  8-inch  wrought-iron  gun  weighing  6£  tons,  and  length 


256 


ELEMENTS  OF  MECHANICS. 


of  bore  126  inches.  The  projectile  was  a  cast-iron  cylinder  15  inches 
long,  turned  to  7.99  inches  in  diameter,  and  weighing  180  pounds. 

The  principal  object  of  these  experiments  was  to  ascertain  the 
maximum  pressure  of,  and  the  effect  produced  by,  different  kinds  of 
ignited  gunpowder  in  the  gun. 

The  powders  experimented  with  were  of  nearly  the  same  compo- 
sition— namely,  saltpetre  75,  charcoal  15,  and  sulphur  10,  making 
100  parts  total. 

Result  of  the  Experiments. 


Nature  of  Ponder. 

Charge  of 
powder. 

Maximum  pres- 
sure per  square 
inch. 

Muzzle  velocity 
per  second. 

Rifle,  large  grain,  R.L.G  

pounds. 
30 

tons. 

29.8 

feet. 
1324 

Russian  prismatic  

32 

20.5 

1366 

Service  pellet  

30 

17.4 

1338 

Pebble  No.  5  

35 

15.4 

1374 

The  rifle  large  grain  is  a  quick  powder,  which  gave  the  smallest 
velocity  with  the  greatest  maximum  pressure. 

The  Pebble,  No.  5,  ia  a  slow  powder,  which  gave  the  greatest  ve- 
locity with  the  smallest  maximum  pressure. 

The  pressure  of  the  ignited  gunpowder  was  measured  by  Rodman's 
pressure-gauge.  The  time  of  operation  in  the  gun  was  recorded  by  a 
chronoscope  designed  by  Captain  Andrew  Noble,  F.  R.  S.,  and  from 
which  the  velocity  of  the  projectile  was  calculated.  (See  London 
Engineer,  Sept.  16,  1870.) 

The  illustration,  Fig.  216,  represents  a  gun  of  the  Rodman  pattern, 
and  not  the  one  used  in  the  English  experiments. 

The  drawn  curves  represent  the  performance  with  powder  of  rifle 
large  grain,  R.  L.  G.,  and  the  dotted  curves  that  of  the  pebble  powder, 
No.  5. 

The  absciss  axis  of  the  diagram  is  divided  into  feet,  and  the  ordi- 
nate  axis  into  pressures  in  tons  per  square  inch. 

The  diagrams  commence  at  the  back  end  of  the  projectile  w  when 
ready  to  fire. 

The  figure  shows  the  position  of  the  projectile  with  35  pounds  of 
pebble  powder.  When  charged  with  30  pounds  of  R.  L.  G.  the  pro- 
jectile is  forced  in  farther  until  it  reaches  the  charge. 


ORDNANCE  DYNAMICS. 


257 


Fig.  216. 


Pressure  Curves. 

The  drawn  line  P' ,  P',  P'  represents  the  pressure  curve  with 
R.  L.  G.,  which  reaches  30  tons  to  the  square  inch.  The  curve  is 
not  continued  to  that  height  on  the  diagram  for  want  of  space,  but 
it  shows  that  that  maximum  pressure  is  instantaneous.  The  ordi- 
nates  measured  from  the  base  line  to  the  curve  show  the  pressure  in 
tons  per  square  inch  at  different  positions  of  the  projectile  until  it 
reaches  the  muzzle. 

The  dotted  line  P,  P,  P  is  the  pressure  curve  for  a  charge  of  35 
pounds  of  pebble  powder. 


258  ELEMENTS  OF  MECHANICS. 

The  pressure  curves  approach  the  shape  of  an  equilateral  hyper- 
bola, but  when  the  projectile  attains  a  greater  velocity  in  the  bore 
the  expansion  of  the  gas  does  not  follow  up  with  its  due  pressure. 

Notation  of  Letters. 

Q  =  weight  in  grains  of  the  powder  in  the  charge. 
q  =  volume  in  cubic  inches  of  the  powder,  including  that,  if  any, 

between  the  charge  and  the  projectile. 
Q  =  volume  in  cubic  inches  of  the  bore. 
q'  =any  volume  of  the  bore  enclosed  by  the  projectile. 

JP  =  pressure  in  pounds  per  square  inch  of  the  ignited  gas. 
/~>i 

—  =  square  of  the  semi-diameter  of  the  hyperbola. 
12 

2.55  is  a  constant  to  be  subtracted  from  the  ordinate  due  to  a 
regular  hyperbola. 

The  formula  for  the  pressure  curve  P'  ,  P'  ,  P'  will  then  be 

P-^-2.55.        .        .        .        .1 

Example.  Charge  of  powder  R.  L.  G.,  30  pounds  x  700  =  210000 
grains  =  G.  Required  the  pressure  of  the  ignited  gunpowder  when 
the  projectile  is  at  the  first  ordinate?  The  projectile  being  moved  in 
3  inches  farther  than  shown  on  the  drawing,  which  position  is  for 
pebble  powder.  The  cross-area  of  the  bore  or  projectile  is  50  square 
inches. 

Volume,  :    27  x  50  =  1350  cubic  inches. 

9  i  noon 

Pressure,  P'  =  -  2.55  =  10.41  tons. 

.  12x1350 

The  dotted  pressure  curve  P,  P,  P  is  for  a  charge  of  35  pounds  of 
pebble  powder,  No.  5,  which  occupies  15  inches  in  the  bore,  or  750 
cubic  inches. 

This  being  a  slow  powder,  drives  the  projectile  ahead  and  increases 
the  volume  before  all  of  it  is  ignited,  and  thus  diminishes  the  max- 
imum pressure,  which  we  find  on  the  diagram  to  be  when  the  projec- 
tile has  moved  about  6  inches,  and  when  the  volume  is  increased  to 
q'  =  50(15  +  6)1050  cubic  inches.  G  =  35  x  7000  =  245000  grains. 


Pressure,         P=   ^          —  2.55  =  16.89  tons. 
The  diagram  shows  only  15.4  tons,  being  slow  powder. 


ORDNANCE  DYNAMICS.  259 

The  treatment  of  pressure,  volume,  temperature  and  work  of  ignited 
gunpowder  belongs  to  dynamics  of  heat,  from  which  the  Formula  1 
is  borrowed. 

Work  Curve. 

The  work  accomplished  by  the  charge  is  represented  by  the  area 
bounded  within  the  pressure  curve  and  abscissa  to  the  ordinate  of  the 
pressure.  This  area,  multiplied  by  the  50  square  inches  section  of  the 
bore,  gives  the  work  in  foot-pounds  accomplished  by  the  ignited 
powder. 

The  work  has  been  calculated  for  each  ordinate,  and  set  off  from 
the  abscissa  to  the  work  curve  K,  K,  K. 

The  number  on  the  scale,  multiplied  by  100,  gives  the  work  accom- 
plished in  foot-tons.  It  is  this  work  which  is  stored  in  the  projectile 
when  set  in  motion  —  namely, 

wV* 
K=  —  —  in  foot-pounds  .....     2 


_ZT=  -  in  foot-tons.       .         .         .         .3 
4480  g 

w  =  180  pounds,  weight  of  the  projectile, 

V—  1374  feet  per  second,  the  muzzle  velocity  of  the  projectile  with 
pebble  powder,  No.  5. 

Work 

The  diagram  shows  2500  foot-tons  accomplished  by  the  charge,  and 
2500  -  2358  =  142  foot-tons,  which  must  have  been  expended  in  fric- 
tion and  leakage  in  the  bore. 

The  diagram  can,  however,  not  be  very  correct. 

Theoretical  Work  in  Ordnance. 

S=  length  in  feet  of  the  bore  of  the  gun. 
s  =  length  occupied  by  the  powder, 
s'  =  length  passed  through  by  the  projectile  in  the  bore. 
A  =  cross-area  of  the  bore  in  square  inches. 
The  differential  work  will  then  be 


12? 
Jr"—  12? 


-2.55        .       ...    4 


260  ELEMENTS  OF  MECHANICS. 

As  the  bore  is  cylindrical,  8s'  =  8q,  and  by  integrating  the  formula 
from  s  to  S,  the  work  will  be 


This  work  should  be  equal  to  that  stored  in  the  projectile  when 
leaving  the  muzzle,  or 


2? 


Having  given  the  charge  G  and  the  dimensions  of  the  gun,  the 
velocity  of  the  projectile  should  be 


The  velocity  will  be  slightly  less  on  account  of  friction  and 
in  the  bore. 

In  a  gun  of  8  inches  in  diameter  of  bore  the  cross-area  is  A  =  50 
square  inches,  and  </  =  12  x  50  =  600  cubic  inches  per  foot  in  the  bore  ; 
then 

1-  and    jf-000,  .....    8 


.9 


This  is  the  formula  for  work  of  gunpowder  in  an.  eight-inch  gun. 

For  the  pebble  powder  we  have  G  =  245000  grains.  The  length 
of  the  gun  8=  10.5  feet,  and  the  length  of  charge  s  =  1.25  feet.  Re- 
quired the  work  of  the  charge  ? 

K=  24o00°  hyp.log.^4-  ~  127.5(10.5  -  1.25)  =  2535  foot-tons. 

The  theoretical  work  is  thus  35  foot-tons  more  than  the  graphical 
work  shown  by  the  diagram,  the  reason  of  which  is  that  the  pressure 
curve  P,  P,  P  does  not  fill  up  the  space  to  the  ordinate  axis,  because 
the  projectile  moved  before  the  due  maximum  pressure  was  reached. 
The  calculation,  however,  proves  the  correctness  of  the  experiments. 


ORDNANCE  DYNAMICS.  261 


Time  Curves. 

The  time  curves  T',  Tf,  T'  for  K.  L.  G.  powder,  and  T,  T,  Tfor 
the  pebble,  No.  5,  were  obtained  by  Noble's  chronoscope,  which  re- 
corded the  moment  the  projectile  passed  each  ordinate.  The  num- 
bers on  the  scale  divided  by  1000  give  the  time  in  decimals  of  a 
second  in  which  the  projectile  reached  each  ordinate. 

The  muzzle  time  for  the  pebble  powder  is  shown  by  the  diagram 

to  be  about  — —  =  0.0106  of  a  second.     The  fifth  ordinate  time  for 
1000 

the  R.  L.  G.  powder  is  — : —  =  0.0055  of  a  second. 
1000 


Velocity  Curves. 

The  velocity  curves  V,  V,  V  for  the  R.  L.  G.  and  F,  V,  V  for 
the  pebble  powder  were  obtained  by  comparing  the  time  and  space. 

Velocity      F=— 10 

The  velocity  of  the  projectile  in  any  part  of  the  bore  can  be  ap- 
proximated by  Formula  7  by  placing  8  and  s'  for  Q  and  q. 


T_  .  ,  s' 

~~  —  12  yp  g'  7  ~       ( 


~  T       ' 


It  is  supposed  in  this  formula  that  the  diameter  of  the  chamber  or 
bore  occupied  by  the  powder  is  equal  to  that  occupied  by  the  pro- 
jectile. 

s  =  length  in  feet  of  the  chamber  or  bore  occupied  by  the  powder. 
s'  =  distance  from  the  bottom  of  the  bore  to  the  projectile  where  the 
velocity  is  required. 

The  starting  velocity  is  slightly  affected  by  different  compositions 
of  powder,  but  the  muzzle  velocity  will  agree  with  the  formula. 

The  number  on  the  scale  of  the  diagram  multiplied  by  100  gives 
the  velocity  in  feet  per  second  of  the  projectile. 

It  will  be  observed  that  the  velocity  curves  tangent  the  abscissas  at 
the  muzzle,  which  proves  that  the  velocity  is  no  more  increased  by 
the  charge,  and  that  the  pressure  of  the  charge  is  reduced  to  almost 
nothing  at  the  muzzle  proves  that  the  gun  is  of  the  proper  length. 


262  ELEMENTS  OF  MECHANICS. 

We  may  estimate  the  proper  length  of  a  gun  as  follows  : 
P  =  -|__2.55=.0,    of  which  ^  =  2.55,    and  «  -  ^  ^  -  JL. 


Q  =  volume  of  the  bore  in  cubic  inches. 
Z)  =  diameter  of  the  bore  in  inches. 
8=  length  of  the  bore  in  feet. 
Q  =  0.942  D*  S. 


Length  of  gun, 


Horse-power   Curve. 

The  horse-power  curve  IT,  JT,  H  is  obtained  by  multiplying  the 
force  or  pressure  by  the  velocity  of  each  ordinate,  and  the  product 

multiplied  by  224°  X  5°  =  203.636. 

550 

The  number  on  the  scale  multiplied  by  100,000  gives  the  acting 
horse-power  at  that  ordinate.  The  maximum  power  for  the  pebble 
powder  is  over  1,900,000  horses. 

The  work  and  horse-power  curves  were  not  given  on  the  English 
diagrams. 

The  six  fundamental  principles  of  dynamics  are  thus  illustrated  in 
the  performance  of  ordnance  —  namely, 


Elements. 
Force          F. 
Velocity      V. 
Time  T. 


Functions. 


Space  S=  V  T. 
Power  F=  F  V. 
Work  K-FVT. 


The  English  experiments  with  different  kinds  of  gunpowder  show 
the  importance  of  applying  the  science  of  dynamics  to  the  perform- 
ance of  heavy  ordnance. 

The  quick  powder,  R.  L.  G.,  gave  an  instantaneous  maximum  pres- 
sure of  nearly  30  tons  to  the  square  inch,  whilst  the  slow  pebble,  No. 
5,  produced  a  continuous  maximum  pressure  of  only  15.4  tons,  and 
gave  a  greater  velocity  to  the  projectile  per  weight  of  powder. 

The  instantaneous  maximum  pressure  is  of  little  or  no  use  in  pro- 
pelling the  projectile,  but  it  injures  the  metal  in  the  chamber  and 
tends  to  burst  the  gun. 

It  appears  that  the  instantaneous  and  excessive  maximum  pressure 
overstrains  the  gas  so  that  it  loses  much  of  its  elasticity,  as  indicated 
by  the  diagram.  The  pressure  above  20  tons  to  the  square  inch 
rises  and  falls  nearly  in  the  same  vertical  line. 


O  UN  PILE-DRIVER. 


263 


Fig.  217. 


The  best  gunpowder  for  heavy  ordnance  is  that  which  gives  the 
greatest  area  of  work  with  the  least  maximum  pressure. 

The  charge  ought  to  be  arranged  with  powder  of  different  quick- 
ness, so  that  a  slow  powder  nearest  to  the  projectile  is  first  ignited, 
than  a  quicker  powder,  until  the  quickest  at  last ;  which  would  gen- 
erate a  low  and  continued  maximum  pressure  with  greater  area  of 
work,  and  which  would  give  much  greater  velocity  of  the  projectile 
with  less  risk  of  bursting  the  gun. 

With  such  arrangement  of  powder  in  the  charge  the  gun  could  be 
made  longer  and  lighter  at  the  bridge. 

The  best  charge  of  powder  is  that  which  gives  the  greatest  radius 
of  curvature  of  the  pressure  curve  at  the  maximum  pressure. 

>,  185.     GUNPOWDER  PILE-DRIVER. 

This  pile-driver,  invented  by  Thomas  Shaw  of  Philadelphia,  is 
worked  by  gunpowder  as  motive-power ;  it 
consists  of  a  gun  a  placed  on  a  pile  b,  and  a 
ram  c  with  a  plunger  d  fitting  closely  in  the 
bore  of  the  gun.  The  ram  is  guided  by  a 
high  framing,  as  represented  by  Fig.  217. 

The  explosion  of  the  gunpowder  in  the  gun, 
drives  the  plunger  d  with  the  ram  c  to  a  con- 
siderable height,  and  the  recoil  of  the  gun 
drives  the  pile  into  the  ground. 

Whilst  the  ram  is  up  a  new  charge  is 
placed  in  the  gun,  and  in  the  fall  of  the  ram 
the  plunger  enters  the  bore  and  compresses 
the  enclosed  air  to  a  high  pressure  and  heat, 
which  ignites  the  charge  for  another  explo- 
sion for  driving  up  the  ram ;  and  so  the  ope- 
ration is,  continued  until  the  pile  is  driven 
home  to  its  destination.  The  pile  is  also 
driven  into  the  ground  by  the  compression 
of  the  air  in  the  gun,  but  the  greatest  drift 
is  in  the  recoil. 

The  ram  can  be  held  at  any  height  by  a 
friction-brake  extending  the  whole  height  of 
the  framing. 

At  the  top  of  the  framing  is  a  cushioning 
piston  e,  which  fits  closely  in  a  bore  in  the 
ram,  for  the  purpose  of  preventing  the  ram  from  striking  or  rising 
above  the  limited  height. 


264  ELEMENTS  OF  MECHANICS. 

The  bores  in  the  gun  and  ram  are  funnel-shaped  at  the  top  for 
admitting  the  plunger  and  piston  with  safety  from  striking  the  edges. 
A  very  slow  gunpowder  should  be  used  in  the  charge. 

I  186.     THEORY  OF  THE  GUNPOWDER  PILE-DRIVER. 

The  recoil  of  the  gun  or  set  of  the  pile  will  be  the  same  as  that  by 
the  Formula  16,  page  253. 


$=set  of  the  pile  in  feet. 

L  =  length  in  feet  of  the  bore  in  the  gun  passed  through  by  the 
plunger,  omitting  the  funnel. 

F=  mean  force  in  pounds  of  the  gunpowder  explosion. 

w  =  weight  in  pounds  of  the  ram  and  plunger. 

W—  weight  in  pounds  of  the  gun  and  pile. 

/=  force  of  resistance  in  pounds  to  the  pile  in  the  ground,  less  the 
weight  W. 

Call  E  =  actual  resistance  to  the  pile,  then  E  =  W+f,  and/=  E  -  W. 

The  efficiency  of  the  pile-driver  consists  in  obtaining  great  recoil, 
and  the  formula  shows  that  the  greater  weight  of  ram  and  the  greater 
length  of  bore  the  greater  will  be  the  set  8  on  recoil ;  but  the  bore 
should  not  be  made  longer  than  is  necessary  for  the  gases  to  raise  the 
ram  a  sufficient  height  for  igniting  the  charge  in  its  fall. 

The  force,  of  gunpowder  cannot  be  correctly  ascertained  without 
experiments,; and  different  compositions  of  powder  give  different  dia- 
grams of  force  and  expansion.  This  subject  belongs  to  dynamics  of 
heat,  from  which  the  following  three  formulas  are  taken.  They  give 
an  approximate  value  of  the  performance  of  gunpowder. 

G  =  weight  in  grains  of  powder  in  the  charge. 

Q  =  volume  in  cubic  inches  of  the  gas  of  the  exploded  gunpowder. 

P  =  pressure  in  pounds  per  square  inch  of  the  gas. 

L  =  length  of  the  bore  of  the  gun  in  feet. 
I  =  length  of  bore  occupied  by  the  charge. 

F=  mean  force  of  the  charge  in  pounds. 

Pressure         P= .  ....     2 

Work  JT-100<?Jj7>.fo$r.(y\        .        .    3 

,    100  a  ,  (L\ 

Mean  force     F= hyp. log.  f  — I       .         .4 

L-l  \  I  j 


GUN  PILE-DRIVER.  265 


Example.  Assume  the  cross-area  of  the  bore  in  the  gun  to  be  40 
square  inches,  and  acted  upon  by  the  explosion  of  O  =  700  grains  of 
powder.  Length  of  the  bore  L  =  2  feet  and  1= 0.08333  of  a  foot, 
which  is  one  inch,  the  distance  between  the  plunger  and  the  bottom 
of  the  gun  at  the  time  of  explosion.  Required  the  pressure  Pand  P' 
per  square  inch  in  the  gun  at  the  time  of  explosion  and  when  the 
piston  leaves  the  muzzle  ? 

The  volumes  of  gas  are  Q  =  1  x40  =  40  cubic  inches  at  the  time  of 
explosion,  and  Q  =  40  x  24  =  960  cubic  inches,  the  volume  of  the  bore. 

100  x  700 
Pressure      P= =  1750  pounds  per  square  inch. 

100x700     , 

Pressure    P  = =  73  pounds  per  square  inch. 

960 

Required  the  work  done  by  the  explosion  ? 
Work     K=  100  x  700  x  hyp.log /— ^— \-  222460  foot-pounds. 

Required  the  mean  pressure  jP? 

K        222460 
Mean  pressure     F=  j—^  =  2_Q()833  =  116066  pounds. 

The  work  K  of  the  explosion  of  the  gunpowder  performs  two 
duties — namely,  to  drive  the  pile  into  the  ground  and  raise  the  ram. 

A  =  the  height  in  feet  to  which  the  ram  is  driven,  the  work  of 
which  is  w  h. 

The  work  of  driving  the  pile  into  the  ground  is  S  f.  Then  we 
have  the  work 

K=wh  +  Sf,        of  which/=— — ,     .        .    5 
and  the  resistance  to  the  pile  in  the  ground  will  be 

R^f+W^^-W.  .    6 

S 

The  bearing  capability  of  the  pile  should  be  calculated  from  For- 
mula 5,  after  the  last  blow  or  set  S. 

The  quality  of  the  gunpowder  can  be  ascertained  experimentally 
by  placing  the  gun  on  a  solid  foundation,  so  that  there  will  be  no 
recoil  of  the  explosion. 

Then,  F(L-t)  =  wh 7 

and  the  mean  force,         F= 8 

L  —  I 

23 


266  ELEMENTS  OF  MECHANICS. 


§137.     COMPRESSION    OF   AIR   IN   THE   GUN. 

When  the  ram  falls  and  drives  the  plunger  into  the  bore  of  the 
gun,  the  air  will  be  compressed  to  a  small  volume  and  heated  to  a 
high  temperature.  Knowing  the  weight  of  the  ram  and  height  of  its 
fall,  it  is  required  to  find  the  compression  and  temperature  of  the  air 
in  the  gun.  For  this  purpose  we  must  borrow  some  formulas  from 
dynamics  of  heat,  which  do  not  belong  to  this  treatise  : 

p  =  pressure  of  the  air  on  the  plunge-piston  when  entering  the 

bore,  which  is  14.7  pounds  to  the  square  inch. 
P=  pressure  of  the  compressed  air  on  the  plunger. 
V=  volume  of  air  of  atmospheric  pressure  in  the  bore. 
v  =  volume  of  the  compressed  air. 


Then,  P=p(  — 


The  bore  of  the  gun  is  cylindrical,  and  the  enclosed  volume  of  air 
is  therefore  as  the  distance  between  the  plunger  and  the  bottom  of 
the  gun. 

Then,  P=P(^T  )"  .....  10 

The  work  accomplished  by  compressing  the  air  will  be 

.      .     .      .11 


This  work  is  wholly  consumed  by  the  generation  of  heat  in  the 
compressed  air.  In  case  the  work  of  the  falling  of  the  ram  w  h  is 

7"  1.4 

greater  than  p  =  —  —  -,  the  difference  will  be  utilized  in  driving  the 

pile  into  the  ground.     This  set  of  the  pile  is  not  included  in  S,  Form- 
ulas 5  and  6. 

P  and  p  mean  the  actual  pressure  above  vacuum  of  the  enclosed 
air  on  the  whole  piston,  from  which  must  be  subtracted  the  atmo- 
spheric pressure  on  the  ram-plunger,  which  is  14.7  A,  when  A  =  area 
in  square  inches  of  the  plunger  piston. 


GUN  PILE-DRIVER.  267 


The  work  done  by  the  ram  in  the  compression  of  the  air  will  then  be 


.14 

and  if  we  include  the  work  that  may  set  the  pile  during  the  compres- 
sion of  the  air,  which  is/s,  we  have 


Theset         s  =  -(w  h-  u.T  A)(-^-+l-L\.     .        .16 

188.     TEMPERATURE   OF   THE   COMPRESSED   AIR   IN   THE   GUN. 

T<=  temperature  of  the  compressed  air. 

t  =  temperature  of  the  air  or  gas  of  atmospheric  pressure  in  the 
gun  when  the  plunger  enters  the  muzzle. 

The  absolute  zero  =  461°  below  zero  of  Fahrenheit's  scale. 

X      /461+JY"  17 


461+JY" 
461  +  *  ) 


161.    .        .        .        .18 

It  is  supposed  in  these  formulas  that  no  air  leaks  out  during  the 
compression,  and  that  no  heat  is  conducted  from  the  air  by  the  metals 
enclosing  it ;  which  cannot  be  the  case,  and  for  which  reason  a  deduc- 
tion of  at  least  25  per  cent,  should  be  made  in  the  length  L. 

The  temperature  of  ignition  of  slow  gunpowder  may  be  assumed 
to  be  T=600°  Fahr.,  and  assuming  the  temperature  of  the  air  or 
gas  in  the  gun  under  atmospheric  pressure  to  be  t  =  80°,  we  have 
461  +  600  =  1061°,  and  461  +  80  =  541°,  the  absolute  temperatures. 

Then 


£  =  5.208  Z,    and    I- 0.192  L, 

when  ignition  of  the  gunpowder  takes  place,  but  if  25  per  cent,  is 
deducted  for  leakage  and  conduction  of  heat,  we  have 

L  =  6.944  I,    and    Z  =  0.144  L. 


268 


ELEMENTS  OF  MECHANICS. 


The  following  data  of  performance  of  the  gunpowder  pile-driver 
has  been  tabulated  by  F.  C.  Prindle,  C.  E.,  U.  S.  N.,  from  records  of 
work  done  at  League  Island  on  the  river  Delaware : 


•s 

8 

Driven. 

Diameter  of  Pile*. 
Inches. 

Distance 
Driren. 

No.  of 

Bloirs  per 

Weight  of 
Powder  per 
Pile. 

5 

| 

1 
* 

| 

Top. 

Bottom. 

Feet 

Pile. 

Pounds. 

II 

=| 

d 

-— 

M 

• 

M 

• 

M 

x 

J 

• 

Tn 

O 

* 

1 

3 

a 

< 

a 

§ 

•< 

3 

a 

< 

a 

a 

< 

a 

a 

< 

fc 

811 

10.5 

13.1 

13 

6 

S.7 

-'2.5 

14 

19.4 

19 

3 

5.2 

1.3 

i 

| 

1300 

6i 

9G6 

15 

10 

12 

11 

6.5 

s 

JO 

21 

24 

... 

20 

... 

2 

1200 

5* 

457 

19 

9 

12.2 

14 

7.5 

8.7 

36 

20 

29.2 

8-5 

11 

H0.4 

U 

u 

3i 

1700 

6f 

63 

16 

10 

12.7 

12 

7.5 

9 

H 

25 

29.5 

122 

39 

59.6 

15 

4 

61 

1200 

5* 

172 

17.5 

9 

11.4 

12 

7 

U 

Hi 

26 

29.2 

30 

6 

12.7 

4} 

1 

3.2 

2170 

7| 

Records  Corresponding  to  the  Numbers  in  the  Table. 

Record  1.  The  first  machine  employed  upon  actual  work.  The 
framing  was  made  of  cast-iron,  mounted  on  a  scow  and  operated  afloat 
at  the  landing  wharf,  Fig.  218.  The  piles  of  heavy  yellow  pine  driven 
through  mud  containing  clay  to  compact  gravel. 

Record  2.  The  framing  of  wood  and  iron,  mounted  on  land.  Piles 
of  hemlock,  and  firmly  driven  without  pointing,  through  stiff  clayey 
material,  mixed  with  sand,  to  hard  gravel  and  boulders.  The  piling 
was  for  the  foundation  for  storehouses,  etc.  Number  of  blows  and 
weight  of  powder  approximate. 

Record  3.  The  same  machine  operating,  in  the  same  kind  of  ground 
as  in  Record  2.  Foundation  for  iron-plating  shop. 

Record  4-  The  same  machine  as  in  Records  2  and  3,  but  with 
lighter  ram. 

Record  5.  The  new  improved  machine,  with  wrought-iron  framing, 
completed  the  work. 


Fig.  218. 


This  illustration,  Fig.  218,  represents  Mr. 
Shaw's  gunpowder  pile-driver  on  a  scow  in  the 
Delaware  River  at  League  Island  Navy- Yard. 


DYNAMOMETER. 


269 


§189.    DYNAMOMETERS. 

Dynamometers  are  instruments  for  measuring  force,  power  and 
work.  The  simplest  form  of  dynamometer  is  that  of  a  spring. 

§190.     SPRING   DYNAMOMETER. 

The  construction  of  this  dynamometer  is  readily  understood  hy 
Fig.  219.  It  is  graduated  by  experiments  in  compressing  the  spring 
with  known  weights. 

Fig.  219. 


This  dynamometer  is,  best  adapted  for  measuring  the  force  of  pull- 
ing a  load  on  a  road,  a  boat  on  a  canal,  or  of  towing  a  ship.  The 
force  in  pounds  indicated  by  the  dynamometer,  multiplied  by  the 
velocity  in  feet  per  second,  will  be  the  power  in  effects,  which  divided 
by  550  will  give  the  horse-power  in  operation. 

2191.     PRONY'S   FRICTION    DYNAMOMETER. 


Fig.  220. 


This  dynamometer  consists  of  a  friction  brake,  as  shown  by  the 
illustration.     It  is  keyed  on  the 
shaft   A,   which   transmits   the 
power  and  work  to  be  measured. 

The  lever  of  the  brake  should 
be  balanced  at  B  before  the 
weight  W  is  put  on  the  scale, 
and  if  it  is  not  balanced,  the 
weight  of  the  lever  and  scale 
should  be  weighed  at  the  scale 
and  added  to  the  weight  W. 

The  weight  W  on  the  scale  is  the  force  acting  on  the  lever  or 
radius  R. 

It  is  supposed  that  all  the  power  and  work  transmitted  by  the 
shaft  is  consumed  by  the  friction  in  the  brake.  When  the  shaft  is 
running  with  its  average  speed  of  n  revolutions  per  minute,  the  strap 
is  tightened  up  with  the  screws,  so  that  the  lever  will  barely  lift  the 
weight  W,  which  is  also  adjusted  to  suit  the  motion.  When  the 
weight  and  friction  are  well  balanced,  count  the  revolutions  per 
minute  of  the  shaft. 

23* 


270  ELEMENTS  OF  MECHANICS. 

The  power  transmitted  through  the  shaft  is  equal  to  the  weight  W 
multiplied  by  the  velocity  of  the  circumference  of  the  radius  It,  mak- 
ing the  same  revolutions  as  the  shaft. 

The  velocity  in  feet  per  second  is 


Power  P  =  W  F=  in  effectef 

which  divided  by  550  give  the 

WEn 


-p.  „         . 

Horse-power     IP  =  —     --  = 

60x550       5253.2 

The  work  Km  foot-pounds  consumed  by  the  friction  of  the  brake 
in  the  time  T  in  seconds  will  be 


Work 


60  9.55 


All  this  work  consumed  by  the  friction  is  restored  by  generating 
heat,  which  makes  the  brake  so  hot  that  a  constant  stream  of  water 
must  run  on  it  to  absorb  the  heat  whilst  the  experiment  is  made, 
otherwise  the  wood  in  the  brake  would  take  fire. 

When  convenient  it  is  best  to  make  the  lever  j£=  10.5  feet,  or  10 
feet  6  inches,  which  will  make  the  circumference  66  feet  ;  in  which 
case,  the  horse-power  will  be 

g^    6&nW  =  2nW 
~  550x60  "    1000  ' 

That  is  to  say,  the  product  of  the  revolutions  per  minute  and 
weight  W,  multiplied  by  2  and  point  off  three  places,  will  be  the 
horse-power  of  the  experiment. 

A  lever  of  R  =  5  feet  3  inches  will  make  the  circumference  33  feet, 
and  the  horse-power 

H.,^. 

1000 


DYNAMOMETER. 


271 


\  192.     BEVEL- WHEEL   DYNAMOMETER. 

Fig.  221  represents  a  side  elevation  and  Fig.  222  a  plan  of  the 
dynamometer. 

Fig.  22L. 


It  can  be  worked  either  by  cranks  c,  c,  or  by  pulleys  a  and  J. 

Let  c?  be  a  shaft  and  pulley  communicating  motion  by  the  dotted 
belt  to  the  shaft  and  pulley  e,  and  it  is  required  to  measure  the  power 
transmitted  between  the  two  shafts.  Place  the  dynamometer  so  that 
it  takes  a  belt/ from  the  pulley  d  on  its  pulley  a,  and  another  belt  g 
from  the  pulley  b  to  the  pulley  e.  The  pulley  a  and  the  bevel-wheel  / 
are  fastened  on  the  crank-shaft. 

The  bevel- wheel  3  is  fast  on  the  pulley  5,  but  both  are  loose  on  the 
crank-shaft. 

The  bevel-wheels  2  and  4  are  both  loose  on  the  arm  h.  Either  one 
of  the  wheels  2  and  4  could  be  dispensed  with,  but  the  dynamometer 
works  better  with  the  two  wheels.  Now  set  the  pulley  and  shaft  c?in 
motion  in  the  direction  of  the  arrow,  and  all  the  power  will  be  trans- 
mitted through  the  dynamometer  to  the  pulley  e.  The  problem  is  to 
measure  the  power  transmitted. 

The  arm  h  is  fitted  loose  on  the  crank-shaft  as  a  fulcrum,  ai'ound 
which  the  arm  is  allowed  a  small  angular  motion,  limited  by  the  slot 
in  the  pillar  i. 


272  ELEMENTS  OF  MECHANICS. 

When  power  is  transmitted  the  bevel-gear  acts  to  lift  the  arm  A, 
but  a  force  F  is  applied  to  keep  the  arm  in  a  horizontal  position.  If 
the  pulley  b  with  the  wheel  3  is  held  stationary,  and  the  arm  be  al- 
lowed to  revolve,  it  would  make  one  revolution  whilst  the  crank- 
shaft makes  two  ;  but  when  the  pulley  b  and  wheel  3  revolve  the  arm 
h  is  held  horizontal  by  a  force  F. 

The  force  F  is  derived  from  a  spiral  spring  enclosed  in  the  drum^, 
which  acts  on  the  chain  like  that  in  a  watch. 

A  ratchet-wheel  is  fastened  on  the  spring  axis,  and  by  the  aid  of 
the  crank  I  the  spring  can  be  regulated  to  the  exact  force  required  to 
balance  the  arm  A  with  the  power  transmitted  through  the  dyna- 
mometer. 

The  drum^  and  ratchet-wheel  k  are  both  graduated  to  indicate  the 
force  F  on  the  chain  in  pounds.  The  sum  of  the  readings  is  the 
force  on  the  chain. 

The  weight  W  is  for  balancing  the  arm  A. 

Notation  of  Letters. 

D  =  diameter  in  feet  of  each  of  the  pulleys  a  and  b. 

f=  force  of  tension  in  pounds  on  the  belts. 
R  =  radius  of  the  arm  A  in  feet. 

F=  force  in  pounds  on  the  chain. 

The  static  momentums  of  the  combination  will  then  be 

F:f=D:R,     and     F  R=f  D. 
FR 


-,     and    / 


When  the  dynamometer  is  worked  by  the  cranks  without  the  belts 
and  pulleys, 

r  =  radius  in  feet  of  the  crank  c. 
f  =  force  in  pounds  on  the  crank-pin. 

F:f'=2r:R,    and     FR-f'Zr. 

2/'r  '  FR 

Jp=__,        and       /=__. 

Transmission  of  Power. 

y=  velocity  in  feet  per  second  of  the  belts. 

v  =  velocity  of  the  end  of  the  arm  A  if  allowed  to  swing. 

n  -=  number  of  revolutions  per  minute  of  the  crankshaft. 


D  YNAMOMETER.  273 


V:v  =  Z>:  E,    and     VE  =  v  D. 

F-—         and  v=  VR 

E  '  D  ' 

T.    -  D  n         ,  -En 

V  = ,    and          v  = . 

60  60 

Power         P  =/  V=  F  v  in  effects. 

Power    fj^.f^. 

The  power  transmitted  through  the  dynamometer  will  then  be 
P  =  *^Ln  =  0.05236  F  E  n. 

Fr.En      FEn 


Horse-power  Jtr  = 


60x550     10503.55 


When  the  dynamometer  is  working  we  have  given  the  force  F  from 
the  spring  graduation.  The  length  E  of  the  arm  h  is  given  and 
constant  for  each  dynamometer.  The  number  of  revolutions  per 
minute  is  obtained  by  counting  the  same.  We  have  thus  given  all 
that  is  necessary  for  calculating  the  power  transmitted  through  the 
dynamometer. 

In  constructing  a  dynamometer  of  this  kind  it  would  be  best  to 
give  such  length  to  the  lever  E  as  to  make  an  even  number  in  the 
denominator  of  the  formula;  for  instance,  if  .#  =  5.251775  feet,  we 
have  the 

Horse-power        IP  =  -  . 
2000 

Allowing  five  per  cent,  for  friction  in  the  dynamometer,  a  five-feet 
lever  would  make  the  formula  the  same  as  above. 

Machinery  very  rarely  transmits  power  uniformly  from  one  locality 
to  another  ;  which  is  particularly  the  case  with  the  ordinary  steam- 
engine,  as  has  already  been  explained. 

The  storage  and  delivery  of  work  by  a  fly-wheel  causes  an  irregu- 
larity in  the  power  transmitted,  which  can  be  measured  by  the 
dynamometer. 

The  oscillation  of  the  arm  h  and  spring-drum  j  indicates  this  irregu- 
larity, which  can  be  read  on  the  graduation,  and  thus  enables  us  to 
determine  with  great  precision  the  efficiency  of  a  fly-wheel.  Suppose 
the  drum  and  ratchet-wheel  to  indicate  a  mean  force  F=  150  pounds 


274 


ELEMENTS  OF  MECHANICS. 


when  the  drum  oscillates  a  difference  12  pounds  ;  the  irregularity  will 
then  be 


,  or  4  per  cent. 


A  dynamometer  of  this  kind  would  be  very  useful  in  institutions 
where  the  subject  of  dynamics  is  taught.  The  students  should  be 
made  to  work  the  cranks  of  the  dynamometer,  and  calculate  their 
own  force,  power  and  work,  and  thus  learn  practically  how  to  dis- 
tinguish the  different  elements  and  functions  in  dynamics,  how  they 
bear  upon  one  another,  and  to  conceive  real  magnitudes  of  such 
quantities. 

The  dynamometer  could  easily  be  arranged  with  indicators  by 
which  to  trace  diagrams  of  the  different  elements  and  functions  in- 
volved in  the  operation. 


§193. 


DYNAMOMETER  AT  THE  ROYAL  TECHNOLOGICAL  INSTITUTE, 
STOCKHOLM. 

Fig.  223. 


The  accompanying  illustration  represents  an  isometric  perspective 
view  of  the  dynamometer  at  the  Royal  Technological  Institute, 
Stockholm.  The  power  is  applied  on  the  crank  a,  and  communicated 
through  the  pulley  b,  rope  V,  pulley  c,  wheel  d,  rope  e,  rope-pulley  g, 
and  is  consumed  by  the  fans///.  The  shafts  n  and  o  are  in  one  line, 
but  not  connected  between  the  pulleys  b  and  c.  The  endless  rope 
V  Vis  held  tight  in  the  grooves  of  the  pulleys  b  and  c  by  means  of 


DYNAMOMETER.  275 


two  weights  w  and  W,  as  will  be  understood  by  the  drawing.  If  the 
two  weights  were  alike,  they  could  communicate  no  motion  to  the 
pulleys,  but  suppose  w  =  10  pounds,  and  W=  20  pounds,  then  there 
would  be  10  pounds  more  weight  on  the  sheave  i,  than  on  the  sheave  k, 
of  which  five  pounds  would  pull  on  each  pulley  b  and  c.  Let  the 
radius  of  the  crank  a  be  equal  to  the  radius  of  the  pulleys,  then  it 
would  require  a  force  of  five  pounds  to  turn  the  crank  in  the  direc- 
tion of  the  arrow.  If  the  crank  is  turned  with  an  irregular  velocity, 
it  would  only  raise  or  lower  the  weights,  but  a  constant  force  of  five 
pounds  would  always  act  on  the  pulley  c  to  communicate  motion  to 
the  fans.  The  power  in  operation  will  be  equal  to  the  force  multi- 
plied by  the  velocity  of  the  rope  F,  and  the  work  accomplished  will 
be  equal  to  the  power  multiplied  by  the  time  of  operation. 

Notation  of  Letters. 

E  =  radius  of  the  crank  in  feet,  which  we  have  supposed  to 
be  equal  to  the  radii  of  the  pulleys  i  and  c. 

F=  force  in  pounds  acting  on  the  crank  a. 

F=  velocity  in  feet  per  second  of  the  rope  V,  which,  in  the 
supposed  case,  will  be  equal  to  that  of  the  crank-pin. 

77=time  of  operation  in  seconds. 
Wand  w  =  weights  on  the  pulleys  b  and  c  in  pounds. 

P=  power  in  dynamic  effects,  of  which  there  are  550  per 
horse-power. 

K=  work,  in  foot-pounds  of  work. 

n  =  number  of  revolutions  per  minute  of  the  pulley  c. 
Then  we  have 

The  force    F=\(W-'w\    and    velocity  F=  2  *  R  ". 
Power         P=FV,     and 


Example.  Radius  of  the  crank  or  pulleys,  .#=1.25  feet,  making 
n  =  28  turns  per  minute.  TF=  80,  and  w  =  20  pounds.  Required,  the 
force  F=?,  velocity  F=?,  power  P=?,  and  how  much  work,  .5T=?, 
will  be  accomplished  in  one  hour,  or  T=  3600  seconds  ? 

Force          F=  |(80  -  20)  =  30  pounds. 

....    2x3.14x1.25x28     _  c.  ,    , 

Velocity      V-  —  —  =  3.66  feet  per  second. 

60 

Power         P=  30  x  3.66  =  109.9  effects. 

Work          K=  109.9  x  3600  =  3956.4  foot-pounds. 


276  ELEMENTS  OF  MECHANICS. 

The  average  power  of  a  man  working  eight  hours  per  day  is  55 
effects,  which  will  be  an  accomplished  work  of  K=  55  x  8  x  3600  = 
1584000  foot-pounds  in  a  day's  work. 

In  order  to  regulate  the  velocity  to  suit  the  power,  the  dyna- 
mometer has  an  arrangement  by  which  to  set  the  fans  at  any  desired 
angle  while  in  motion,  which  arrangement  is  not  shown  on  the  draw- 
ing. Students  used  to  work  the  dynamometer  in  a  spirit  of  emula- 
tion to  outdo  each  other  in  power  and  work.  Some  could  accomplish 
the  greatest  power,  and  work  with  less  force  and  more  velocity, 
whilst  others  preferred  more  force  and  less  velocity. 

Arrangements  could  easily  be  made  to  register  on  the  dynamometer 
the  force,  velocity,  power,  and  work  in  the  form'of  diagrams. 

Dynamometers  of  this  kind  ought  to  be  employed  in  all  scientific 
institutions  where  dynamics  are  taught,  for  we  have  yet  no  better 
means  by  which  to  imbue  the  student  with  the  real  substance  of  dy- 
namics. Any  student  who  has  worked  this  instrument  for  a  few 
hours  will  probably  not  commit  the  error  of  saying  that  work  is  in- 
dependent of  time,  or  that  time  is  included  in  power,  which  erroneous 
ideas  are  yet  maintained  in  text-books. 

This  dynamometer  does  not  only  teach  the  student  the  different 
properties  of  force,  power  and  work,  but  it  enables  him  to  conceive  and 
compare,  with  great  precision,  real  magnitudes  of  those  quantities, 
which  is  of  great  importance  in  designing  machinery. 

In  the  year  1850  the  author  made  a  great  many  experiments  with 
different  kinds  of  screw-propellers,  in  which  was  employed  a  dyna- 
mometer of  this  description,  made  by  Thomas  Mason  of  Philadelphia, 
and  which  gave  great  satisfaction ;  and  he  has  always  considered  it 
the  best  form  of  dynamometer  where  it  can  be  conveniently  applied. 


SOUND.  277 


§  194     DYNAMICS   OF   SOUND. 

Sound  is  work,  consisting  of  the  three  simple  elements  F  V  T,  of 
which  F=  force  of  the  sound,  F=  velocity  of  vibration,  and  J"=time 
of  continuance  of  the  sound. 

The  loudness  of  sound  is 

Power         P=FV. 

The  pitch  of  the  sound  indicates  the  proportion  between  F  and  V. 

Two  different  sounds  of  different  pitch  may  be  of  equal  power  or 
loudness,  but  the  high-pitch  sound  is  then  produced  by  small  force  F 
and  great  velocity  F",  whilst  the  low-pitch  sound  is  produced  by 
greater  force  F'  and  smaller  velocity  V  ',  so  that  the  products  F  V 
and  F'  V  are  alike  in  the  two  sounds. 

Let  an  elastic  spring  a  b  be  drawn  aside  a  space  S  where  the  force 
is  F.  Leave  the  spring  to  take  its  own  course,  f.  224 

and  it  will  move  fore  and  back  and  set  the 
surrounding  air  into  vibration,  which  produces 
sound. 

The  work  expended  in  drawing  the  spring 
aside  is  K=F  S,  which  work  is  restored  by 
producing  sound,  and  the  spring  will  continue 
to  sound  until  all  the  work  F  Sis  consumed. 

The  loudness  or  power  P=  F  Vof  the  sound 
is  greatest  at  the  start  of  vibration,  after  which 
it  will  gradually  diminish,  and  finally  fade 
away  to  nothing. 

Differential  work  file  =  Pfit,  but  the  power  decreases  as  the  time  in- 

G 
creases  that  we  can  place  P  =  —  ,  in  which  C  is  a  constant  factor. 


,     Ctit        T^    rCdt 
£  =  — ,       K=  \ = 

t  J     t 


T\      F  9 

Time  of  sound,  hyp.log.  T=  —  =  -  -. 
0         0 

The  force  F  of  an  elastic  spring  is  as  the  space  8  within  the  limit 
of  vibration,  and  the  mean  force  in  the  space  8  is  therefore  \  F.  The 
spring  vibrates  the  same  pitch  of  sound  in  any  space  8,  and  the  pro- 
portion between  ^"and  Vis  therefore  constant.  The  loudness  of  the 
sound  or  power  jP=  F  V=  OS'1,  or  the  power  of  the  sound,  is  as  the 
.square  of  the  space  of  vibration.  The  velocity  V  in  the  above  for- 
mulas means  that  of  the  force  producing  the  sound,  and  not  the  ve- 
locity of  the  sound  from  the  sonorous  body. 
24 


278  ELEMENTS  OF  MECHANICS. 

Noise  in  Machinery. 

All  sound  or  noise  in  operating  machinery  represents  so  much  work 
lost,  and  the  loudness  of  the  noise  represents  the  power  lost.  The 
noise  in  a  cotton-mill,  rolling-mill,  or  railroads,  etc.  represents  power 
and  work  lost. 

Noise  of  a  Steam-hammer. 

A  steam-hammer  falling  on  its  bare  anvil  will  set  the  whole  system, 
with  the  surrounding  air,  into  vibration,  and  create  a  great  noise, 
which  represents  the  whole  work  of  the  hammer ;  but  the  same  ham- 
mer falling  upon  a  puddle-ball,  or  upon  white-hot  iron,  will  create 
very  little  noise,  and  the  greatest  part  of  the  work  of  the  hammer  is 
then  utilized  in  forging  the  iron. 

Report  of  a  G-un. 

The  report  of  a  gun  represents  the  work  lost  of  the  total  work  due 
from  the  ignited  powder.  If  the  ignited  gas  in  the  gun  could  be 
allowed  to  propel  the  projectile  until  its  force  of  expansion  is  reduced 
to  the  atmospheric  pressure,  there  would  be  no  loud  report 

Sound  of  a  Bell. 

The  work  of  a  clapper  in  striking  its  bell  is  represented  by  the 
sound  produced.  M  =  mass  of  the  clapper  and  F=  velocity  of  the 
strike.  Then  the 

MV* 
Work  of  the  clapper  = =  work  of  the  sound. 

Sound  of  Men  and  Animals. 

The  sound  of  men  or  animals  is  produced  by  the  heat  in  the  body, 
which  is  work.  A  speaker,  singer,  or  a  musician  playing  a  wind  in- 
strument draws  from  his  store  of  heat  as  when  doing  hand  labor,  and 
the  louder  he  speaks,  sing?,  or  plays,  the  greater  power  is  drawn  from 
his  heat,  and  he  ultimately  becomes  fatigued  as  when  working  a  crank. 

Music  of  an  Organ. 

The  weight  on  the  organ-bellows  multiplied  by  the  velocity  with 
which  it  sinks  whilst  air  is  discharged  and  none  enters  is  the  power  in 
effects,  which,  divided  by  550,  gives  the  horse-power  of  the  organ. 
The  same  weight  multiplied  by  the  space  it  sinks  is  the  work  done  by 
the  organ  in  producing  the  music. 

Dynamics  of  Sound  includes  the  science  of  acoustics,  and  is  a 
very  extensive  and  interesting  subject,  which  requires  a  separate 
treatise. 


SOUND. 


279 


g  195.     VELOCITY    OF   SOUND    IN   AIR. 

The  velocity  of  sound  in  air  Las  been  determined  both  by  experi- 
ments and  theory  by  Newton,  Laplace,  Dalton  and  various  others; 
the  summary  of  which  is 


V=  1089.42^7 1  +  0.00208(!!  -  32). 


D  =  1089.42  TI/  1  +  0.00208(2  -  32). 

V=  velocity  in  feet  per  second  of  the  sound. 
t  =  temperature  Fahr.  of  the  air. 

D  =  distance  in  feet  travelled  in  the  time  T  in  seconds. 
1089.42  =  velocity  of  sound  at  32°,  temperature  of  the  air. 
0.00208  =  volume  of  expansion  per  degree  Fahr.  of  the  air. 


Distance  in  Feet  which  Sound  Travels  in  Air  at  Different 
Temperatures 


Time 
see. 

, 

TEMPERA' 
10°  |  20° 

CURE  0 

32° 

F  THE 

40° 

AIR, 
50° 

FAHRI 
60° 

INHEH 

70° 

P  SCAI 

80° 

-E. 
^ 

100° 

1 

1000 

1064.2 

1075.7 

1089.4 

1098.5 

1109 

1120 

1131 

1142 

1153 

1164 

2 

1985  i  2128 

2151 

2179 

2197 

2219 

2241 

2262 

2285 

2306 

2328 

3 

2978 

3193 

3227 

3268 

3295 

3328 

3361 

3393 

3427 

3459 

3492 

4 

3971 

4257 

4303 

4358 

4394 

4438  |  4482 

4524 

4570 

4613 

4656 

5 

4964 

5321 

5378 

5447 

5492 

5548 

5603 

5655 

5712 

5766 

5821 

6 

5956 

6385 

6454 

6536 

6591 

6657 

6723 

6786 

6855 

6919 

6984 

7 

6949 

7449 

7530 

7626 

7689 

7767 

7844 

7917 

7997 

8072 

8148 

8 

7962 

8514 

8606 

8715 

8788 

8876 

8964 

9049 

9140 

9225 

9312 

9 

!  8934 

9578 

9681 

9805 

9886 

9986 

10085 

10180  10282 

10379 

10476 

10 

!  9927  10642 

10757 

10894 

10985  11096  11306 

11311  11425 

11532 

11640 

11 

110920  11706 

11833 

11983  12083112205 

12326 

12442  12567 

12685 

12804 

12 

11912  12770 

12908 

13073 

13182  13315  13447 

1357313710 

13838 

13968 

13 

12905  13835 

13984 

14162 

14280  14424 

145(57 

14704  14852 

14991 

15132 

14 

13898  14899 

15060 

15252 

15379 

15534 

15688 

15835 

15995116145 

16296 

15 

14891  15963 

16135 

16341 

16477  116644 

16809 

169661  17137 

17298 

17460 

16 

!l5883  i  17027 

17211 

17430 

17576  17753 

17929 

18097118280 

18451 

18624 

17 

16876  17091 

18287 

18520 

186741886319050 

19228!  19422!  19604 

19788 

\  18 

1788919156 

19363 

19609 

19773  19972 

20170 

20360  20565 

20757 

20952 

19 
20 

18861  20220 
19854  21284 

20438 
21514 

20699 
21788 

20871  2108221291 
21970  !22192i22412 

21491  21707  21911 
22622  122850  23064 

22116 
23280 

21 

'20847  '22348 

22590 

22877 

23063  23301 

23532 

23753  1  23992 

24217 

24444 

22 

21839  23412 

23665 

23967 

24167  124411:24653 

24884  25135  25376  1  25608 

23 

22832  24477 

24741 

25056 

25265  2552025773 

26015  26277 

26523  26772 

24 

23825  25541 

25817 

26146 

26364 

26630  26894 

27146  27420  27677/27926 

25 

24818  26605 

26892 

27235 

27462 

27740 

28015 

28277  28562 

28830  29100 

280  ELEMENTS  OF  MECHANICS. 

§  196.    COUNTING  BEATS  OF  SECONDS. 

When  the  occurrence  of  a  distant  sound  is  not  anticipated,  we  are 
unprepared  to  record  the  exact  moment,  and  before  an  appropriate 
timekeeper  can  be  procured  an  uncertain  time  has  elapsed. 

With  some  practice  the  beats  of  seconds  can  be  counted  in  the  mind 
with  tolerable  correctness  without  the  aid  of  a  timekeeper;  which 
practice  has  been  of  great  service  to  the  author  in  astronomical  obser- 
vations. Practice  to  count  seconds  by  the  aid  of  an  oscillating  second 
pendulum,  or  by  the  second-hand  on  a  watch,  until  the  counting  agree 
with  the  timekeeper,  without  attention  to  the  pendulum  or  second- 
hand. With  good  practice  the  counting  should  not  differ  more  than 
one  second  per  minute. 

When  an  unexpected  distant  sound  is  heard  and  its  cause  observed, 
we  can  always  be  ready  to  count  seconds,  and  thus  determine  the 
distance. 

In  astronomical  observations  at  sea  it  is  customary  to  keep  a  watch 
in  the  hand,  or  to  station  an  assistant  at  the  chronometer  to  note  the 
time  when  the  observer  says  "  stop  ;"  but  there  are  known  cases  when 
the  captain  has  taken  his  observations  without  the  aid  of  a  watch  or 
assistant,  and  walked  slowly  and  comfortably  to  his  cabin  and  noted 
the  time  of  his  observations  from  the  chronometer,  with  no  little 
amusement  to  other  observers,  who  naturally  supposed  that  the  cap- 
tain's observations  could  not  be  very  correct,  but  to  their  surprise 
were  found  to  be  as  correct  as  their  observations  with  ordinary  precau- 
tions. The  captain  counted  in  his  mind  the  beats  of  seconds,  and  de- 
ducted the  sum  from  the  time  observed  on  the  chronometer. 

The  practice  of  counting  seconds  correctly  is  of  great  utility  and 
service  for  estimating  various  movements.  When  the  action  is  of 
very  short  duration,  say  less  than  3  seconds,  it  is  best  to  count  half 
seconds,  or  even  four  times  per  second,  and  a  short  time  may  be  de- 
termined with  a  correctness  within  a  quarter  of  a  second. 


ASTRONOMY.  281 


ASTRONOMY. 

CREATION  OF  WORLDS. 

§  197.  MATTER  in  celestial  space  arranges  itself  into  groups  or 
nebulas  by  virtue  of  universal  attraction,  which  by  the  aid  of  centri- 
fugal force  are  finally  divided  into  definite  bodies,  of  which  the  largest 
occupies  the  centre  around  which  the  smaller  revolve,  and  the  group 
is  called  a  planetary  system. 

Each  fixed  star  is  a  central  body  of  a  planetary  system  like  our  sun. 

The  rotary  motion  of  each  group,  nebula  or  of  each  body  around  its 
axis  has  been  caused  by  collisions  of  the  matters  constituting  that 
group  or  body. 

The  conditions  under  which  a  nebula  can  be  formed  into  a  plan- 
etary system  are — -first,  that  it  must  be  set  into  a  quick  rotation; 
secondly,  that  it  must  consist  of  different  kinds  of  matter  ;  and  thirdly, 
that  its  ingredients  must  be  of  such  proportions  as  to  admit  of  divis- 
ion by  the  forces  of  attraction  and  centrifugal. 

Without  the  above  conditions  the  nebula  will  remain  permanent 
until  it  comes  into  collision  with  some  other  nebula  or  body. 

It  appears  that  each  kind  of  matter  is  derived  from  different  parts 
of  space,  and  in  its  course  of  travel  meets  and  mingles  with  other 
kinds  of  matter,  whereby  nebulas  of  a  variety  of  shapes  are  formed. 
The  different  forms  of  nebulas  are  caused  by  different  dynamical  and 
chemical  actions  of  the  mingling  matters.  The  act  of  collision  of 
nebulas  is  distinctly  seen  through  powerful  telescopes,  and  the  mag- 
nitude of  the  collision  is  so  enormous  that  a  change  in  form  is  hardly 
perceptible  during  a  lifetime  of  observations.  The  forms  of  the  dif- 
ferent nebulas  indicate  their  relative  ages,  which  may  be  graduated 
from  the  form  of  a  group  of  clouds  to  that  of  a  permanently  organized 
planetary  system. 

The  majority  of  the  well-defined  nebulas  located  within  our  scope  of 
observation  are  spherical  or  egg-shaped,  with  a  bright  central  spot 
indicating  the  act  of  forming  a  sun,  about  which  the  surrounding 
matter  gradually  divides  itself  into  separate  masses,  forming  at  length 
a  planetary  system. 

Some  nebulas  are  of  the  form  of  a  ring,  others  consist  of  one  or 
more  spirals — a  configuration  which  indicates  the  relative  motions  of 


282  ELEMENTS  OF  MECHANICS. 

its  constituent  elements.  We  also  find  nebulas  of  the  form  of  a 
spindle,  having  a  bright  body  at  each  end. 

The  irregular  nebulas,  which  are  of  the  form  of  a  group  of  clouds, 
may  be  classed  as  primary  formations. 

The  period  of  time  in  which  a  primary  nebula  is  thus  formed  into 
a  planetary  system  may  be  many  millions  of  years. 

The  operation  of  forming  nebulas  and  the  creation  of  worlds  is 
going  on  all  around  us,  even  within  our  limit  of  observation  through 
powerful  telescopes,  but  the  magnitude  of  that  operation  is  too  enor- 
mous for  any  human  mind  to  conceive.  The  work  is  gradual  and 
continued  until  all  the  matter  in  each  part  of  space  has  assumed  a 
definite  form  and  motion.  The  matter  in  our  part  of  space — that 
is,  the  space  occupied  by  our  planetary  system  and  the  neighboring 
ones — seems  to  be  arranged  into  a  definite  form  and  motion ;  but  in 
other  parts  of  space,  where  groups  and  bodies  are  yet  forming,  some 
matter  or  portions  of  bodies  may  be  led  astray  by  repulsive  force  in 
collisions,  and  by  being  overtaken  by  superior  force  of  attraction  from 
other  systems  is  drawn  toward  a  central  body,  as  is  the  case  with 
matter  constantly  flowing  into  our  sun. 

When  such  stray  matter  is  in  the  form  of  a  solid  body  of  perhaps 
thousands  or  millions  of  cubic  miles  in  volume,  it  makes  spots  in  the 
sun's  photosphere ;  but  a  great  deal  of  such  flowing  matter  is  in  the 
form  of  a  gas  or  powder  (which  we  sometimes  see  as  zodiacal  light 
when  it  passes  near  to  the  earth)  which  makes  no  visible  spots  in  the 
sun.  Stray  matter  is  often  taken  up  by  planets,  as  experienced  on 
our  earth,  arid  we  call  it  meteors. 

Falling  meteors  change  the  motion  of  the  earth  and  disturb  our 
chronology,  but  their  mass  is  so  very  small  compared  with  that  of  the 
earth  that  it  requires  many  years  of  observation  for  us  to  appreciate 
any  such  change ;  and  as  the  meteors  fall  from  all  possible  directions, 
some  of  them  may  counteract  the  action  of  their  predecessors.  The 
meteors  which  have  fallen  on  the  earth  within  our  time  of  records  and 
tradition  have  not  changed  our  chronology  more  than,  perhaps,  a  few 
seconds ;  but  there  has  evidently  been  a  time  when  very  large  bodies 
have  struck  the  earth  and  changed  its  rotation  both  around  its  axis 
and  around  the  sun,  before  which  time  the  present  location  of  the 
poles  might  have  been  at  or  near  the  equator. 

The  largest  known  meteor  on  our  earth  is  lying  on  the  pampa  of 
Tucmnan,  near  Otumpa,  in  the  Republic  of  Argentine,  South  America, 
weighing  about  16  tons. 

A  Comet  is  stray  matter  seeking  a  situation  to  supply  other 
heavenly  bodies  with  oxygen,  hydrogen  and  nitrogen,  and  travels 


CREATION  OF  WORLDS.  283 

from  system  to  system,  describing  elliptical  orbits  around  each  central 
body,  until  its  course  is  by  chance  directed  near  enough  to  some 
planet  or  sun  to  catch  and  retain  the  comet. 

The  return  of  a  comet  cannot  be  calculated  or  predicted  except 
when  its  orbit  is  limited  within  our  system.  When  the  orbit  extends 
outside  of  our  system,  the  comet  will  likely  never  return,  but  is  over- 
taken alternately  by  other  planetary  systems  or  groups  of  matter. 

A  planetary  system  may  also  have  two  suns,  and  subdivisions  of 
groups  or  nebulas,  in  which  the  revolving  bodies  are  called  satellites, 
moving  around  a  planet  and  forming  a  system  within  itself,  but 
subject  as  one  body  to  the  main  system.  There  are  several  planets 
in  our  system  forming  such  a  group,  of  which  the  earth  and  moon  are 
one.  The  planets  Saturn  and  Uranus  have  each  eight  satellites, 
Jupiter  has  four,  and  Neptune  one. 

The  condition  under  which  the  revolving  bodies  are  maintained  in 
their  regular  orbits  is,  that  the  force  of  attraction  of  the  central  body 
is  equal  to  the  centrifugal  force  of  the  revolving  one. 

The  orbits  of  periodical  rotation  are  ellipses,  in  which  the  central 
body  is  in  one  of  the  foci.  The  orbit  may  accidentally  be  a  circle, 
but  there  is  no  known  planet  or  satellite  which  revolves  in  a  perfect 
circle.  A  pendulum  freely  suspended  and  made  to  swing  so  as  to 
describe  a  cone,  the  base  of  that  cone  would  practically  be  an  ellipse, 
for  the  reason  that  it  is  almost  impossible  to  start  the  pendulum  with 
such  perfect  velocity  and  direction  in  relation  to  its  radius  as  to  make 
it  swing  a  perfect  circle.  Such  is  the  case  with  the  planetary  orbits, 
in  which  the  planets  have  not  been  started  so  as  to  describe  perfect 
circles,  and  the  orbits  are  also  disturbed  by  the  attraction  between 
the  planets. 

2198.     OUR    PLANETARY   SYSTEM. 

It  will  be  observed  in  the  following  tables  that  the  other  worlds 
in  our  system  are  of  a  different  nature  from  that  of  our  earth,  and 
that  no  two  of  them  are  alike.  Their  difference  of  density  is  a  strik- 
ing feature ;  some  of  them  are  light  as  wood,  others  heavy  as  rock ; 
and  the  planet  Mercury  must  evidently  be  a  chunk  of  precious  metals. 
The  density  of  the  sun  is  given  to  be  1.128,  compared  with  that  of 
water,  but  the  photosphere  is  included  in  the  volume  of  that  mean 
density.  The  core  of  the  sun  is  evidently  a  very  dense  body.  The 
column  of  density  of  the  planets,  however,  indicates  the  tendency  of 
the  heavy  materials  to  occupy  the  inner,  and  the  lighter  the  outer, 
portion  of  a  nebula  or  planetary  system. 

The  following  tables  contain  the  principal  elements  of  our  planetary 
system : 


284 


ELEMENTS  OF  MECHANICS. 


Elements  of 

the  Planetary 

System. 

The 

• 

Mean  distance  from 

Sidereal  peri 

od  of     |  Rotat'n 

Eccentric  of 

principal 

I 

the  sun. 

one  revolution.        !  ar.  axis. 

orb.  in  part  of. 

planets. 

55 

Earth  =1;          Miles. 

Days. 

Yrs.  mU. 

Hours. 

Sem.  axis. 

Sun 

60748 

Mercury 

6 

0.3871 

36,774,000 

87.96926 

0    2.93 

24.05 

0.2056179 

Venus    . 

9 

0.7233 

68,613,000 

224.7008 

0    7.49 

23.21 

0.0068334 

Earth     . 

© 

1.000000 

95,000,000 

365.25637 

1       00 

24.00 

0.01677046 

Moon 

c 

Frt>m  ©  '         237,360 

27.32116 

0    0.91 

0.0 

0.0635 

Mars      . 

<? 

1.5236 

144,742,000 

686.97964 

1 

10.72 

24.37 

0.0932616 

Jupiter  . 

% 

5.2028 

494,266,000 

4332.5848 

11 

10.49 

9.56 

0.0482388 

Saturn    . 

h 

9.5388 

906,186,000 

10759.220 

29    5.56 

10.29 

0.0559956 

Uranus  . 

tf 

19.182 

1822,290,000 

30686.820 

84       3 

9.30 

0.0465775 

Neptune 

V 

30.037 

2853,515,000 

60126.72 

165     7 

.      . 

0.0087195 

The 

* 

Diameter 

Velocity  in 

Surface 

Attraction 

Light  and 

Sizeofsun 

principal 

fl 

in 

orbit  m 

ties 

Off 

lanet. 

on  s 

urface. 

ht.  fr.  s 

in. 

seen  from  | 

planets. 

7. 

miles. 

per  secc 

nd. 

Earth  =  1 

Earth  =  1 

Earth  =  1 

planet. 

Suh 

882000 

1243 

28.3 

Mercury 

0 

5 

3140 

30.4 

0.144 

0.51 

6.673 

2.584 

Venus    . 

7800 

22.3 

0.973 

0.91 

1.910 

1.383 

Earth     . 

© 

7912 

18.9 

1.000 

1.00 

1.000 

1.00 

Moon 

C 

2160 

6.35 

0.0746 

0.167 

1.000 

1.00 

Mare      . 

j 

4100 

11.33 

0.269 

0.500 

0.430 

0.656 

Jupiter  . 

V 

87000 

8.31 

121 

2.456 

0.037 

0.192 

Saturn    . 

h 

79160 

6.14 

80 

1.09 

0.011 

0.105 

Uranus  . 

ft 

34500 

4.33 

19 

1.05 

0.003 

0.052 

Neptune 

V 

41500 

3.45 

27.6 

1.10 

0.001 

0.033 

The 

B 

Mass  of 

Volume 

Density  of  the 

Substances 

Inclination  of 

principal 

planet. 

of  planet. 

planets. 

of  equal 

o 

rbit  to 

planets. 

8 

Earth  =  1 

Earth  =  1 

Earth  =  1 

Water 

=  1 

gravity. 

ecliptic. 

Sun    .     . 

0 

355000 

1378000 

0.2543 

1.128 

Boxwood  . 

Mercury 

9 

0.6966 

0.06218 

2.782 

11.6 

Lead     .     . 

7°    0'   8".  16 

Venns    . 

9 

0.877 

0.9531 

0.9434 

4.19 

Topaz  .     . 

3°  23'  30".  75 

Earth     . 

9 

1.000 

1.0000 

1.0000 

5.44 

Iron  pyrites 

0 

0       0 

Moon      . 

d 

0.0114 

0.02024 

0.615 

3.345 

Limestone 

5 

3   8 

'48". 

Mars      . 

tf 

0.1313 

0.1384 

0.1293 

0.575 

Pine      .     . 

1 

3  51 

'   5".  08 

Jupiter  . 

^ 

317.5 

1322.5 

0.2589 

1.15 

Oak       .     . 

1 

5  18 

'40".  31 

Saturn    . 

h 

139.5 

996.2 

0.1016 

0.451 

Charcoal   . 

2°  29'  28".  14 

Uranus  . 

y 

198. 

82.47 

0.2796 

1.24 

Bit  u  in.  coal 

0°46 

'  29".  91 

Neptune 

V 

20. 

143.5 

0.222 

0.988 

Camphor  . 

1 

'  46'  58".  97 

INHABITATION  OF  WORLDS.  285 


§  199.    INHABITATION    OF   WORLDS. 

It  would  be  unreasonable  to  suppose  that  our  little  earth  is  the 
only  inhabited  world  amongst  the  millions  of  worlds  in  the  universe, 
for  wherever  the  proper  proportions  of  the  elements  of  life  exist,  or- 
ganic bodies  are  formed  in  accordance  with  the  conditions  of  operating 
elements. 

Different  organic  bodies  are  composed  of  different  material  elements 
and  ingredients  in  different  proportions.  The  principal  material  ele- 
ments required  for  the  support  of  life  are  oxygen,  hydrogen,  nitrogen 
and  carbon,  and  the  physical  elements  are  force,  motion  and  time, 
which  constitute  the  functions  light  and  heat. 

We  know  the  operation,  distribution  and  proportions  of  these  ele- 
ments on  the  earth's  surface,  but  have  no  correct  knowledge  of  the 
same  in  other  worlds. 

In  regard  to  the  other  worlds  in  our  system  we  know  that  the  in- 
tensity of  light  on  them  is  inversely  as  the  square  of  their  distance* 
from  the  sun,  but  that  does  not  give  the  heat  and  climate  on  those 
worlds.  The  heat  which  supports  organic  life  on  our  earth  is  gen- 
erated by  the  light  passing  through  our  atmosphere,  and  the  denser 
the  air  is  the  more  heat  is  generated,  as  proven  by  the  decrease  of 
temperature  with  the  altitude  above  the  level  of  the  sea.  The  per- 
petual snow-line  in  the  tropics  is  about  15,200  feet  above  the  level 
of  the  sea,  which  proves  a  temperature  of  32°  Fahr.  at  that  height. 

Heat  is  generated  by  light  passing  through  any  transparent  me- 
dium, and  when  the  light  is  weak  a  denser  medium  is  required  for 
generating  the  heat  necessary  for  the  support  of  organic  life. 

As  an  illustration  of  the  generation  of  heat  by  light,  suppose  a 
hollow  cylinder  or  tube  of  say  3  feet  long  and  6  inches  inside  diameter, 
with  a  bottom  in  one  end,  made  of  a  non-conducting  substance  for 
heat.  Insert  about  forty  pieces  of  plate-glass  cut  to  fit  the  inside 
circle  of  the  tube,  which  will  be  about  three-fourths  of  an  inch  be- 
tween each  plate,  leaving  two  or  three  inches  between  the  last  plate 
and  the  bottom.  Place  the  tube  so  that  the  sun's  light  passes  straight 
through  it,  and  a  heat  will  be  generated  at  the  bottom  far  above  the 
temperature  of  boiling  water. 

The  reason  of  this  is,  that  light  is  power,  which  is  composed  of  force 
and  velocity ;  and  when  the  velocity  is  reduced  by  the  glass  plates, 
the  force,  which  is  temperature,  will  be  increased. 


286  ELEMENTS  OF  MECHANICS. 

MOON. 

The  nearest  world  to  our  earth  is  the  moon,  which  we  know  has 
little  or  no  atmosphere,  and  is  therefore  not  likely  to  be  inhabited, 
although  she  receives  the  same  intensity  of  light  from  the  sun  as  does 
the  earth.  The  topography  of  the  moon's  surface  is  very  clear  through 
powerful  telescopes,  but  no  sign  of  habitation  has  yet  been  discovered 
there.  The  planets  are  too  remote  for  minute  examination  by  the 
limited  power  of  our  present  telescopes,  but  they  are  no  doubt  more 
or  less  inhabited. 

MERCURY. 

The  planet  Mercury  is  nearest  to  the  sun,  and  receives  over  six 
times  the  intensity  of  light  as  does  the  earth,  but  with  a  light  atmo- 
sphere, inhabitation  is  possible  if  the  other  elements  necessary  for  the 
support  of  life  exist  there.  Mercury  is  an  irregularly  shaped  mass 
of  precious  metals,  and  cannot  possess  the  abundance  of  the  more  use- 
ful elements  composing  our  earth. 

VENUS. 

Venus  is  the  second  planet  from  the  sun,  and  is  nearly  of  the  same 
size  and  composition  as  our  earth,  but  the  intensity  of  her  light  is 
about  double  that  at  the  earth.  The  planet  Venus  is  no  doubt  well 
inhabited.  It  is  the  nearest  planet  to  us,  but  being  inside  of  the 
earth's  orbit,  the  sun's  light  interferes  with  our  telescopic  views  of  her 
topography. 

EARTH. 

The  third  planet  from  the  sun  is  the  Earth,  upon  which  we  live 
amidst  its  abounding  glories  established  for  us  by  the  Creator  of  the 
Universe.  Its  population  is  about  1,400,000,000  inhabitants,  which  is 
the  only  known  datum  in  the  table  of  population  of  our  planetary 
system. 

We  have  also  thousands  of  different  species  of  animals,  insects  and 
plants  to  make  up  the  inhabitation  of  our  world. 

MARS. 

Mars  is  the  fourth  planet  from  the  sun,  and  receives  0.43  the  in- 
tensity of  light  as  does  the  earth.  The  surface  of  this  planet  has  a 
very  conspicuous  appearance,  indicating  land  and  seas,  with  a  dense 
atmosphere  in  which  we  see  floating  clouds,  and  is  probably  inhabited 
with  organizations  suitable  to  the  conditions  of  its  elements. 

Mars  is  the  first  planet  upon  which  we  expect  to  discover  inhabita- 
tion when  our  telescopes  are  sufficiently  advanced  for  that  purpose. 


PLANETS.  287 


ASTEROIDS. 

Between  the  orbits  of  Mars  and  Jupiter  are  a  number  of  planetoids, 
which  are  probably  destined  to  be  consolidated  into  one  body  with 
satellites.  About  120  of  them  have  been  discovered  and  named,  of 
which  Ceres,  Pallas,  Juno,  Vesta,  Astraea,  Hebe,  Iris  and  Flora  are  the 
principal  ones. 

These  planetoids  are  called  Asteroids,  and  are  probably  not  in- 
habited, on  account  of  not  having  been  long  enough  in  a  stable  or 
permanent  condition.  Some  low  grade  of  organizations  may  exist  on 
some  of  them. 

JUPITER. 

This  is  the  largest  planet  in  our  system,  and  counting  the  asteroids 
as  one,  Jupiter  is  the  sixth  planet  from  the  sun.  The  intensity  of  his 
light  from  the  sun  is  only  0.037  of  that  of  our  earth,  but  he  is  sur- 
rounded with  a  dense  atmosphere,  in  which  can  be  seen  floating  clouds. 

Jupiter  has  four  satellites,  forming  a  complete  system  within  itself. 
The  surface  of  Jupiter  indicates  the  existence  of  land  and  seas  well 
denned,  and  is  probably  inhabited. 

SATURN. 

Saturn  is  the  seventh  planet  from  the  sun,  and  is  surrounded  with 
concentric  rings,  which  appear  in  small  telescopes  to  be  only  one  ring. 
This  planet  has  eight  satellites,  forming  a  system  within  itself. 

The  intensity  of  light  in  Saturn  is  only  0.011  of  that  on  our' earth. 
Saturn  is  too  far  from  the  earth  for  the  limited  power  of  our  present 
telescopes  to  examine  its  surface,  but  is  probably  inhabited. 

URANUS. 

Uranus  is  the  eighth  planet  from  the  sun,  and  is  accompanied  with 
a  number  of  small  satellites,  of  which  eight  have  been  discovered. 

Satellites  generally  revolve  around  their  planet  in  the  same  direc- 
tion as  the  planet  around  the  sun,  and  in  orbits  of  a  small  inclination 
to  the  plane  of  the  planet's  orbit ;  but  the  satellites  of  Uranus  re- 
volve in  an  opposite  direction,  and  with  an  inclination  nearly  at  right 
angles  to  the  plane  of  the  planet's  orbit. 

NEPTUNE. 

Neptune  is  the  last  known  planet  in  our  system,  and  was  discovered 
by  Leverrier  in  the  year  1846 ;  it  is  not  visible  to  the  naked  eye,  and 
can  be  observed  only  through  a  powerful  telescope,  by  which  one     • 
satellite  has  been  found  to  accompany  the  planet. 


288  ELEMENTS  OF  MECHANICS. 

SUMMARY   OF   INHABITATION   AND   CIVILIZATION    IN   WORLDS. 

Inhabitants  of  other  worlds  are  as  comfortable  with  their  combi- 
nation of  material  and  physical  elements  as  we  are  with  ours.  They 
have  different  lengths  of  years,  seasons,  nights  and  days;  different 
force  of  attraction,  atmospheric  pressure,  light  and  heat,  as  shown  in 
the  table  of  elements  of  our  planetary  system.  A  body  weighing  one 
pound  on  the  earth's  surface  weighs  2.456  pounds  on  Jupiter,  and 
only  half  a  pound  on  the  planet  Mars.  The  light  on  the  surface  of 
Uranus  is  only  0.003  of  that  on  the  Earth,  but  the  optical  organs  of 
its  inhabitants  (if  such  exist)  are  constructed  accordingly,  so  as  to 
render  them  as  comfortable  with  their  light  as  we  with  ours. 

Each  variety  of  inhabitants  is  necessarily  accommodated  to  the 
conditions  of  the  operating  elements,  the  most  perfect  organizations 
requiring  more  complicated  elementary  combinations. 

We  can  justly  claim  that  man  is  the  most  perfect  organization 
known  on  our  earth,  but  that  claim  cannot  be  extended  to  other 
worlds,  particularly  as  long  as  we  maintain  armies  and  navies  for 
offensive  and  defensive  purposes ;  whilst  the  various  forms  of  mischief, 
egoism  and  malignity  which  exist  among  men  on  earth,  do  not  speak 
well  for  their  civilization. 

Considering  the  progress  of  man  within  the  scope  of  history,  and 
the  fact  that  a  large  portion  of  the  human  race  is  yet  in  its  primitive 
state,  totally  estranged  from  the  surrounding  progress,  it  appears  that 
our  earth  cannot  be  very  old  in  its  present  permanent  condition.  The 
idea  of  age  in  this  connection  comprehends  millions  of  years. 

We  have  good  reason  for  supposing  that  organizations  in  other 
worlds  outside  of  our  system  are  far  superior  to  our  own,  because  the 
character  of  organizations  does  not  depend  only  upon  the  existence 
of  all  the  material  and  physical  elements,  but  principally  upon  their 
proportions  and  distribution,  which  are  evidently  better  classified  in 
other  worlds  or  in  permanent  nebulas  existing  millions  of  years  before 
our  earth  was  formed.  The  superior  organizations  in  such  old  worlds 
have  advantages  not  only  in  time  and  experience,  but  in  greater 
varieties  of  physical  phenomena  upon  which  to  exercise  their  know- 
ledge and  intelligence. 

They  may  be  sufficiently  advanced  in  the  science  of  optics  to  be 
able  to  extend  their  vision  to  our  earth  and  examine  our  doings. 

We  can  therefore  be  convinced  that  there  exist  in  other  worlds 
beings  which  are  far  superior  to  ourselves,  whilst  above  all  presides 
the  Creator  of  the  Universe,  who  superintends  these  myriad  organi- 
zations, whose  infinite  inventions  testify  to  His  exhaustless  and  eternal 
power. 


ASTRONOMY.  289 


§200.    LAW  OF  CELESTIAL  MECHANICS. 

The  law  of  celestial  mechanics  was  partly  anticipated  by  several 
savans  before  and  in  Newton's  time,  but  they  did  not  succeed  in 
arranging  the  physical  elements  so  that  the  combinations  would  agree 
with  their  thoughts  and  observations. 

Newton  received  a  very  valuable  assistance  from  Kepler,  as  may 
be  inferred  from  the  correspondence  between  the  two  savans ;  which 
correspondence  most  likely  led  to  the  final  establishment  of  the  laws 
of  celestial  mechanics  by  Newton,  which  are  as  follows : 

1st.  The  areas  described  by  the  radius-vector  of  a  planet  is  equal  in 
equal  times. 

2d.  The  planetary  orbits  are  ellipses,  in  which  the  sun  is  in  one  of 
the  foci. 

3d.  The  force  of  attraction  between  any  two  masses  is  as  the  product 
of  the  masses  and  inversely  as  the  square  of  their  distance  apart. 

4th.  The  square  of  the  times  of  one  revolution  of  the  planets  are  as 
the  cubes  of  the  semi-major  axis  of  the  orbits. 

In  the  accompanying  illustration  the  ellipse  m,  a,  b,  c  represents 
the  orbit  of  a  planet  m  revolving  around 
the  sun  M.  R=*  radius-vector,  or  distance 
between  M  and  m.  T=  time  of  one  rev- 
olution, and  V=  velocity  of  the  planet  in  the 
orbit. 

In  accordance  with  the  first  law,  the  area 
described  in  a  unit  of  time  by  the  radius- 
vector,  say  from  m  to  a,  is  equal  to  the  area 
described  in  the  same  unit  of  time  in  any  other  part  of  the  ellipse, 
say  from  b  to  c,  or  the  areas  of  the  sectors  m  Ma  =  b  Me. 

The  second  law  defines  the  orbit  in,  a,b,cto  be  an  ellipse  in  which 
the  sun  Mis  in  one  of  the  foci. 

The  third  law  defines  the  force  of  attraction     F= . 

jR 

The  centrifugal  force  of  the  planet  m  is  F=         •. 

M 

m  V1    Mm         ,     ,.,.    M  r,       IM 

Therefore     _  _- — ,    and     F'  =  -,    or     F-^-. 

As  the  mass  M  of  the  sun  is  constant  for  all  his  planets,  it  follows 
that  the  velocity  Fin  the  orbit  is  inversely  as  the  square  root  of  the 
radius-vector. 

25  T 


290  ELEMENTS  OF  MECHANICS. 

The  semi-major  axis  op  is  a  function  of  the  time  T  of  one  revolu- 
tion of  the  planet  in  its  orbit,  but  op  is  a  function  of  the  radius- 
vector  R,  for  which  reason  we  can  place 

R-VT,    of  which     F=|,    and     F'-f-f. 

The  mass  M  is  constant  for  all  the  planets,  and  we  have 

T*  -  R,  the  fourth  law. 

The  square  of  the  times  of  one  revolution  are  as  the  cubes  of  the 
radius-vector. 

I  201.    TO  FIND  THE  MASS  OF  THE  SUN. 

The  centrifugal  force  of  the  earth  revolving  around  the  sun  is  equal 
to  the  force  of  attraction  between  the  two  bodies. 
M=*  mass  of  the  sun,  and  in  =  mass  of  the  earth. 
D  =  distance  in  feet     J  from  ^  earth  to  ^  ^ 
R  =  distance  in  miles  j 
n  =  revolutions  per  minute  of  the  earth  around  the  sun. 

Centrifugal  force  F°= . 

TT  i     •,.  rr    ^TtDn          , 

Velocity  V= ,   and 

60 

Centrifugal  force  J?=  — 


Z>60* 

Centrifugal  force,  m  lA-—\  =  ^          force  of  attraction. 

\    oU   /      ^jooyoUou  JJ' 

/9  TT  W\2 

Mass  of  the  sun,  M  =  28693080 1?t J ,  in  matts. 

The  sidereal  number  of  revolutions  per  minute  of  the  earth  around 
the  sun  will  be 

1  =      1 

60x24x365.25     511350' 

log.b.  7087 182. 
2_  1 

"259679000000' 

%.11.4174364. 


%.5.9190422. 


ASTRONOMY.  291 


D  =  5280  E  =  5280  x  95,000,000  =  501,600,000,000  feet, 
I?  =  126,203,844,096,000,000,000,000,000,000,000,000  cubic  feet. 
From  paragraph  51  we  know  that  the  mass  of  the  earth  is 

402,735,000,000,000,000,000,000  matte.  ^.23.6050086. 
Then  the  mass  of  the  sun  compared  with  that  of  the  earth  will  be 
126,203,844,096,000,000 
"    402,735^831002" 

%.5.5764616. 

This  result  is  a  little  higher  than  that  in  the  table,  page  284. 

|  202.     TO   FIND   THE  DISTANCE  FROM  THE  SUN  TO  ANY  OF  HIS 
PLANETS. 

Knowing  that  the  centrifugal  force  of  any  planet  revolving  around 
the  sun  is 

FWi 

and  that  this  centrifugal  force  is  equal  to  the  force  of  attraction, 


we  have 


Z  a  nV     ,,  „     Ml  60   * 

* 


TV  4.  n      'W  60  V 

Distance        D 


Let  t  denote  the  time  in  minutes  of  one  sidereal  revolution  of  any 
planet  around  the  sun. 

Then  t  •=  -,  and  11  =  -,  which  inserted  in  the  formula  for  D,  we  have 
n  t 


292  ELEMENTS  OF  MECHANICS. 

Let  T  denote  the  time  in  days  of  one  sidereal  revolution  of  a  planet 
around  the  sun,  and  we  have 

*=rx  60x24  =  1440  T. 


n      »/  JS//86400  T 

^  \7V~2T- 

In  this  formula  we  have  given  the  masses  of  the  sun  : 

Log.  mass  of  the  earth 23.6050086 

Log.  earth  mass  of  sun add.     5.5800176 

29.1850262 


Log.  <p sub.    7.4577772 


Log.  — 21.7272490 


add.     8.2771082 


Divide  by  3 30.0043572 

Log.  10033500000  = 10.0014524 


D  =  10033500000^  T.  "* 
Let  R  denote  the  radius  of  the  orbit  of  a  planet  in  statute  miles. 


1900290.       %6.2788185. 


Radius-vector     E  =  1900290^1^ 

This  formula  gives  the  distance  from  the  sun  to  any  of  his  planets, 
when  the  time  Tin  solar  days  of  one  sidereal  revolution  is  known. 

Example.  '  It  is  known  that  the  planet  Jupiter  makes  one  revolu- 
tion around  the  sun  in  a  time  T=  4332.6  days.  Required  his  mean 
distance  from  the  sun  ? 


E  =  1900290^/4332.6*  =  505030000  miles. 

There  is  one  item  omitted  in  the  sections  201  and  202 — namely,  that 
the  planets  do  not  revolve  strictly  around  the  sun's  centre,  but  around 
the  common  centre  of  gravity  of  the  two  bodies.  The  whole  plan- 
etary system  revolves  around  its  common  centre  of  gravity. 


ASTRONOMY.  293 


|203.     TO   FIND   THE   TIME   IN    WHICH   TWO   BODIES  WOULD   BE 
DRAWN  TOGETHER  BY  THE  FORCE  OF  THEIR  OWN  ATTRACTION. 

Assume  two  bodies  M  and  m  to  be  held  at  a  distance  S  apart  ;  when 
let  loose  or  free  to  move,  their  force  of  attraction  will  draw  them 
together. 

Fig.  226. 


—  ^.-TxRi-r*.  ------  n  --------  .> 

Let  Jfand  m  denote  the  respective  masses,  expressed  in  matts. 
8=  distance  apart  in  feet. 

T=  time  in  seconds  in  which  they  would  be  drawn  together. 
"Fand  v  =  velocities  in  feet  per  second. 
R  and  r  =  the  respective  radii  of  the  masses  in  feet,  supposing  them 

to  be  spherical. 
a  and  b  =  the  respective  distances  in  feet  moved  by  the  masses  to 

the  point  of  collision. 
<P  =  28693080,  the  coefficient  of  attraction. 
As  the  force  of  attraction  is  the  same  on  each  body,  we  have 

--    m      a  in  b 

a:b  =  m:M,    —  =7,     or    a  =  ^rr. 
M     o  M 


M 


"Ft 
F:m=v:t,    of  which    v  = . 


Mmt  8b 


ft  &  = 

J 


25  • 


294  ELEMENTS  OF  MECHANICS. 

Example.  Suppose  the  moon  to  be  stopped  in  her  circular  motion 
around  the  earth ;  the  force  of  attraction  would  then  draw  the  two 
bodies  together. 

It  is  required  to  know  the  time  in  which  the  moon  would  fall  to  the 
earth  ? 

The  masses  of  the  earth  and  moon  are, 

M-  402, 735,000,000,000,000,000,000  matts...%  23.6050194 

w  =  4,591, 000,000,000,000,000,000 log.  21.6619135 

#=1253260800  feet '......log.    9.0980413 

,S  =  20887680  feet,    and    r  =  5702400  feet. 

b  =  121282000  feet log.     8.0838052 

?  =  28693080 log.     7.4577772 

— -0.0114,         /l-  —  \  =0.9886 log.    0.9941411-1 

M  \       Mi 

Insert  these  values  in  Formula  1,  and  the  result  will  be  171180 
seconds,  which  is  1  day,  23  hours  and  33  minutes,  the  time  required 
for  the  moon  to  fall  upon  the  earth. 

A  very  small  body  m  falling  from  a  great  distance  S  into  a  very 

large  body  M,  makes  [1 J  =  1,    and    b  =  S,  nearly  ;  and  the  time 

of  fall  may  be  estimated  by 
t 


The  bodies  M  and  m  are  moved  in  opposite  directions  with  their 
common  force  of  attraction ;  the  velocities  Fand  v  will  consequently  be 
V:v  =  m:M,    and    MV=mv. 

The  momentums  being  alike,  the  masses  will  stop  one  another's 
motion. 

Supposing  the  bodies  to  have  no  elasticity  and  to  be  crushed  into 
one  another  in  the  collision,  the  work  of  that  collision  will  be  the 
sum  of  the  works  stored  in  each  body. 

TIT  ~rr* 
The   work  in   M  is   JfT= 

The    work    in    m    is    1c  = 


2 
m  v3 


•nr     i       f        iv   •  TJT     i 

Work  of  collision     K +  k  = 


ASTRONOMY.  295 


The  consolidated  masses  will  be  brought  to  rest  after  the  work  of 
collision. 

The  work  stored  in  each  mass  is  inversely  as  the  masses   and 
velocities. 

Mmfia 


The  force  of  attraction  between  the  earth  and  moon  is 
Mm      402,735etc.  x  4,591etc. 


<p  &     28693080  x  1253260800s 


=  41027200000000000000  pounds. 


If  this  force  was  constant  and  applied  on  the  moon  in  the  direction 
of  the  earth,  it  would  bring  the  two  bodies  together  in  a  time 

„       12  MS       I  2x4591etc.xl253etc.       ,onfl1A 
2*—  ^j  --  =  A  —  -  -  529610  seconds, 

\     F       \  410272etc. 

or  6  days,  3  hours,  6  minutes  and  45  seconds. 

§204.     MAXIMUM   ATTRACTION. 

The  force  of  attraction  between  any  two  bodies  M  and  m  at  a  dis- 
tance S  aart  is 


f  jfi? 

The  force  of  attraction  is  inversely  as  the  square  of  the  distance, 
consequently,  when  the  distance  is  infinitely  small,  the  attraction  will 
be  infinitely  great  ;  but  there  is  a  limit  to  the  distance  between  the 
centres  of  attraction  —  namely,  when  the  bodies  are  in  close  contact 
their  centres  of  attraction  are  still  at  a  considerable  distance  apart. 

Let  two  spherical  masses  M  and  m,  and  of  radii  R  and  r,  be  in 
close  contact  to  one  another  ;  their  maximum  attraction         ~    ^ 
will  then  be 

Jfm 


If  the  masses  are  forced  into  one  another,  the  force  of  attraction 
will  be  diminished. 


296  ELEMENTS  OF  MECHANICS. 

The  masses  of  spheres  of  equal  substance  is  as  the  cube  of  the  radii, 
and  the  force  of  attraction  can  be  expressed  by 


. 

(S+rf 

When  the  two  spheres  are  of  equal  radii,  or  E  =  r,  the  force  of 
attraction  is 

jyjy_jy 

' 


D  =  2  jft,  the  diameters  of  the  sphere,  and  the  force  of  maximum 
attraction  is  a  limited  function  F=  Z>*. 

That  is  to  say,  the  force  of  attraction  between  equal  spheres  in  con- 
tact is  as  the  fourth  power  of  the  diameter,  or  the  force  of  attraction 
increases  directly  as  the  fourth  power  of  the  distance  between  the 
centres  of  attraction. 

§205.     ELEMENTS   OF   THE    EARTH   AND   MOON. 

The  earth  and  the  moon  revolve  around  their  common  centre  of 
gravity  in  periods  of  29.53  days  from  new  moon  to  new  moon. 

M  =  mass  of  the  earth,  and  m  =  that  of  the  moon. 
r  =  radius  of  the  earth,  and  JR  =  distance  from  the  centre  of  the 
earth  to  the  centre  of  the  moon.     It  =  60  r. 

Let  c  denote  the  position  of  the  common  centre  of  gravity  of  the 
two  bodies,  a  =  distance  from  c  to  the  centre  of  the  earth,  and  b  =  dis- 
tance from  c  to  the  centre  of  the  moon. 

Let  d  denote  the  position  of  the  centre  of  attraction  of  the  earth 
and  moon.  It  is  required  to  find  the  centres  c  and  d? 

Centre  of  Gravity  of  the  Earth  and  Moon. 

In  regard  to  the  centre  of  gravity  c,  we  have  the  static  momentums 


m  b     ra(60  r  -  a) 


'-©-^ 

»        I 

tf*    !  a  =  - 

}\    Jt  MM 

^fcv!  JHfa  =  w60r-wa; 

Jfa+ma  =  60wr,     or    a(  Jf +  m)  =  60  m  r. 
60mr 
Jf+m 


ASTRONOMY.  297 


Assume  the  radius  of  the  earth  r  =  1. 

Log.60      =   1.7781513 
kg.m        =  21.6619135 
=  23.4400648  + 
=  23.6050086- 


%.0.684  =   9.8350562 

That  is,  the  distance  of  the  centre  of  rotation  of  the  earth  and  moon 
is  at  0.684  r  from  the  centre  of  the  earth,  or  0.316  of  the  earth's  radius 
under  the  surface  of  the  earth  where  the  moon  is  in  the  zenith. 

The  centres  of  the  earth  and  moon  thus  describe  elliptic  orbits 
around  the  centre  c  as  the  common  focus,  whilst  the  centre  c  moves 
in  an  elliptic  orbit  around  the  sun. 

The  radius  of  the  earth  is  about  3956  miles,  or  20887680  feet. 

a  =  3956  x  0.684  =  2705.904  miles. 

b  =  3956  x  60  -  2705.904  =  157534.096  miles. 

The  velocity  of  a  body  moving  in  a  circle  is  -  . 

The  earth  and  moon  make  one  sidereal  revolution  in  27.32  solar 
days.  27.32  x  24  =  655.68  hours. 

2x3.1472         R 


Velocity, 


655.68       104.407' 


•n    ,1,  2705.904     nK  Mt,     ., 

Earth  s  centre,        v  -  —      —  —  25.917  miles  per  hour. 
104.407 

Moon's  centre,         "F= — — — —  =  1508.8  miles  per  hour. 

104.407 


Centre  of  Attraction  Between  the  Earth  and  Moon. 

In  regard  to  the  centre  of  attraction  between  the  earth  and  moon, 
we  have 

Force  attraction,          F= . 

I? 

It  is  required  to  find  a  point  in  the  straight  line  between  the  earth 
and  the  moon  where  a  body  would  be  equally  attracted  from  the  two 
bodies,  which  is  when 

m- ,        of  which  d= 


d*    (R-d?  '    M-m> 


298 


ELEMENTS  OF  MECHANICS. 


Assuming  the  mass  of  the  moon  as  unit,  or  m  =  1,  then  the  mass 
of  the  earth  will  be  M  =  87.7.  When  E  =  60  the  distance  d  will  be 
expressed  in  radii  of  the  earth. 

[)  =  54.2. 


d=  54.2x3956  =  214415  miles  from  the  centre  of  the  earth  to  the 
centre  of  attraction.  A  body  placed  at  e,  Fig.  228,  would  be  equally 
attracted  from  the  earth  and  moon. 


|206.     ORBITS   OF   THE   EARTH   AND   MOON. 

Fig.  229  represents  the  orbits  of  the  earth  and  moon  moving  in  the 
direction  of  the  arrow,  as  seen  from  the  North  Star.  The  dotted  line 
a  represents  the  orbit  described  by  the  common  centre  of  gravity  of 
the  earth  and  moon,  which  is  an  ellipse  in  which  the  sun  is  in  one 
of  the  focii.  The  eccentricity  of  the  ellipse  is  only  0.0168  of  the 
major  axis. 

Fig.  229. 


The  drawn  line  b  represents  the  orbit  described  by  the  centre  of 
the  moon,  and  the  line  c  that  of  the  earth.  The  line  0, 1,  2,  3  and  8  re- 
presents the  radius-vectors  from  the  sun.  The  illustration  represents 
the  orbits  for  one  lunar  month,  or  from  new  moon  to  new  moon,  which 
are  accomplished  in  a  time  of  29  days,  31  minutes  and  48  seconds. 

The  mean  velocity  of  the  common  centre  of  gravity  in  the  dotted 
orbit  a  is  68,091  miles  per  hour. 

The  following  table  shows  the  mean  sidereal  velocities  in  miles  per 
hour  of  the  earth  and  moon  in  their  respective  orbits  at  different 
positions  of  the  moon. 


ASTRONOMY. 


299 


Sidereal  Velocities  of  the  Earth  and  Moon. 


No. 

Positions. 

Velocities  in  n 
Earth. 

ilcs  per  hoar. 
Moon. 

0 
1 
2 
3 

4 
5 
6 

7 
8 

68086 
68110 
68091 
68072 
68069 
68072 
68091 
68110 
68086 

66582 
67024 
68091 
69158 
69600 
69158 
68091 
67024 
66582 

Middle  of  first  quarter 

Middle  of  second  quarter  

Full  moon  

Middle  of  third  quarter  

Half  moon  

Middle  of  fourth  quarter  

New  moon  

The  greatest  difference  of  velocity  of  the  moon  is  3013  miles  per 
hour,  a  distance  of  about  that  between  New  York  and  Liverpool. 

The  moon  travels  a  distance  of  about  10  times  the  diameter  of 
the  earth  per  hour. 

A  point  on  the  equator  describes  a  worm-line  on  the  earth's  orbit. 

§  207.     TO  FIND  THE  MASS  OF  THE  EARTH  AND  MOON. 

The  notation  of  letters  will  be  the  same  as  in  the  preceding  para- 
graph. 

The  earth  M  and  moon  m  revolve  around  their  com-       Fi&-  m 
mon  centre  of  gravity  c.     Having  given  the  earth's  radius 
and  the  moon's  horizontal  parallax,  we  obtain  the  distance 
R  -  237360  miles,  or  P.  =  1253260800  feet,  between  the 
centres  of  the  two  bodies. 

The  force  of  attraction  between  the  two  bodies  is 


F= 


in  which  y  =  28693080. 


(2  7T    \*  / 

ecTr)  =mb( 


T=time   in   minutes  of  one   sidereal   revolution   of  the   system, 
which  is 

27*,  7*,  43",  IP,     or     T=  39343.183  minutes. 


Call    »-I-^ 


2x3.14 


60x39343.183 


141150000000 
log. 1 1.1496796. 


300  ELEMENTS  OF  MECHANICS. 

The  centrifugal  forces  are  equal  to  the  force  of  attraction,  or 

Mm      ,..    .  ,  Mm         ,  „ 

and—  -»i». 


, 

m  .     J/       , 

—  —  =  a  o,  and  -  =  00. 

<p  IP  y  R* 

28693080  1 

Call    Z=  <p  IS  - 


141150000000     4919.26 

%.3.691919024. 


The  mass  of  the  earth  and  moon  M+m  =  - 
12532608003 


4919.26     • 
M+m  =  402,820,000,000,000,000,000,000  matte. 

This  calculation  gives  the  mass  of  the  earth  and  moon  a  little  less 
than  that  on  page  294,  owing  to  the  data  not  being  very  correct. 

The  masses  of  the  earth  and  moon  cannot  be  calculated  separately 
from  the  preceding  formulas,  because  the  given  data  R  and  T  are 
constant  for  any  proportion  of  M  and  m.  If  M  =  m,  the  centre  c 
would  be  in  the  middle  between  the  earth  and  moon,  but  R  and  T 
would  be  unaltered,  and  the  force  of  attraction  and  centrifugal  forces 
would  still  be  alike. 


g  208.      TO   FIND   THE   MASSES   OF  ANY   TWO   HEAVENLY   BODIES 
REVOLVING   AROUND  ONE  ANOTHER. 

Let  8  denote  the  distance  in  statute  miles  between  the  centres  of 
the  two  bodies,  and  £  =  time  in  days  of  one  sidereal  revolution,  J/and 
m  being  the  masses  of  the  revolving  bodies  expressed  in  units  of  that 
of  the  earth  and  moon. 


17914680000000  f 

fo$r.!3.2532071. 


ASTRONOMY.  301 


Required  the  mass  of  the  earth  and  the  sun  ? 

8  =  95,000,000  miles,        t  =  365.242  days. 

17914680000000  x  365.25637  4°'5' 

When  the  mass  of  the  earth  is  1,  that  of  the  sun  will  be  358739.5. 
When  the  mass  of  one  of  the  bodies  is  known,  that  of  the  other  is 

*'          -jr. 


17914680000000  ? 

The  mean  distance  of  the  planet  Jupiter  from  the  sun  is  estimated 
to  be  £=494265000  miles,  and  the  planet  makes  one  sidereal  revo- 
lution in  t  =  4332.6  days.  Required  the  mass  of  Jupiter  ? 


17914680000000  x  4332.6" 


Log.£=   8.6939599 

mult,  cube 3 

26.0818797 

subt.  20.5267043 

%359067=   5.5551754 


log.t-   3.6367486 

mult,  square 2 

7.2734972 

log. 119,  etc.  - 13.2532071 
20.5267043 


The  masses      M+m  =  359067 
Mass  of  the  sun     M=  358739 


Mass  of  Jupiter     m  =       328 

This  includes  also  the  mass  of  Jupiter's  satellites. 

The  mass  can  thus  be  calculated  of  any  planet  whose  distance  from 
the  sun  and  time  of  sidereal  revolution  is  known. 

When  the  sum  of  the  masses  of  two  revolving  bodies  is  known, 
their  distances  apart  S  in  miles  and  time  t  in  days  of  one  sidereal 
revolution  will  be  as  follows : 


Distance    8=  26165.9f/  f(M+m)  in  minutes. 
Time  t  =  4232563-^-^  in  days. 


302  ELEMENTS  OF  MECHANICS. 


Volume  and  Density. 

For  the  volume  and  density  of  the  heavenly  bodies  it  is  necessary 
to  know  their  mass  and  diameter. 

D  =  diameter  of  the  planet  in  statute  miles. 
P  —  volume  in  cubic  miles. 
Q  =  density  compared  with  water  at  39°  Fahr. 
M=  mass  of  the  planet  compared  with  that  of  the  earth. 

Volume    P 
Density     Q 


. 

6 

1400350000000  M 


. 

Example.     Required  the  volume  and  density  of  the  earth  ? 
D  =  7912  miles,  the  diameter  of  the  earth. 

Volume    P=  3.14x7912*  =  ^WUWWQQ  cubic  miles. 
6 

1400350000000 
259014000000 

That  is  to  say,  the  earth  is  5.4025  times  heavier  than  an  equal 
volume  of  water. 

The  interior  of  the  earth  is  probably  composed  of  metallic  sulphurets, 
principally  iron  and  copper.  The  specific  gravity  of  these  substances 
is  about  5.4. 

The  specific  gravity  of  the  planet  Mercury  is  variously  stated  be- 
tween 6  and  15  times  that  of  water  ;  it  must  evidently  be  a  purely 
metallic  body. 

The  volume  and  density  of  the  planet  Jupiter  will  be,  when 
Z>  =  87000,  and  Jf=328. 

Volume     P-3'14x8700°3  -  336,770,000,000,000  cubic  miles. 
6 

1400850000000x828 


336770000000000 

The  density  of  Jupiter  is  1.36389  compared  with  that  of  water,  or 
0.252  of  that  of  the  earth. 

The  astronomical  data  appear  to  be  yet  very  incorrect,  and  it  is 
therefore  difficult  to  make  them  agree  with  the  physical  laws  involved. 


ASTRONOMY.  303 


g  209.     TO   FIND   THE   VELOCITY   WITH   WHICH   A    BODY   SHOULD 
be  started  from  the  earth's  surface  in  order  to  reach  the  Moon. 

Fig.  231. 

In  the  accompanying  illustration  M  represents  the  earth 
and  m,  the  moon.  The  point  m'  is  the  centre  of  attraction 
between  the  two  bodies.  A  body  at  m'  would  be  equally 
attracted  by  M  and  m  ;  therefore  if  the  body  m'  is  overtaken 
by  the  earth's  attraction,  it  would  arrive  at  b  with  a  velocity 
equal  to  that  with  which  it  should  be  started  from  the 
earth's  surface  in  order  to  reach  the  moon. 

M,  m,  and  m'  denote  the  masses  of  the  respective  bodies 
expressed  in  matte. 

S=  distance  in  feet  between  the  centres  of  the  earth  and  moon. 
d  =  distance  in  feet  of  the  body  m'  from  the  centre  of  the  earth. 
F=  force  of  attraction  of  the  earth  on  the  body  m'. 
/=  force  of  attraction  of  the  moon. 

_    M  m          ,      .       mm 

,   and   f-. 


The  force  acting  on  the  body  m'  in  its  fall  to  the  earth  will  be 
_     .. 


-  f=  -  /— -      m      \ 

J  v\#   (£-<*)*/ 


Let  v  denote  the  velocity  of  the  fall  in  feet  per  second. 
77=  time  of  fall  in  seconds. 

(F-f)  :M=v:T,    of  which     (F-f)  -  ~. 

m!  v    m'  I M         m 
~T~~ 


_T_lM__m_\  ^ 


304  ELEMENTS  OF  MECHANICS. 

ri  T 

T=-  —  -,  when  d  and  v  are  variables,  which  inserted  in  Formula  1 


ill  be 


8d  /M 

v  --  /  --- 


/n         1    rlMfia         mda    \ 
v  8v  =  —  I) i. 
<pJ\   d*        (8-d?) 

*.    ±JX ™\  +  c 

2      j\d     (S-d)) 


The  operation  is  to  be  integrated  from  the  centre  of  attraction  to 
the  surface  of  the  earth,  where  d=R  the  radius  of  the  earth,  and 
when  v  =  0  we  have 


=     JL/"^        m      \ 
9\d    (S-d)J' 


Then          V-M± ™ &+-OL 


9       \Rd    (&. 
We  have  all  the  data  given  in  this  formula — namely, 

M-  402,735,000,000,000,000,000,000  matte 23.6050194 

m  =  4,591,065,000,000,000,000,000  matts 21.6619135 

(2=214415  miles  =  1132111200  feet 9.0538890 

£=1253260800  feet 9.0980413 

£  =  20887680  feet 7.3198903 

?  =  28693080 7.4577772 

The  body  m'  will  then  fall  on  the  earth  with  a  velocity  F=  40781 
feet  per  second. 

That  is  to  say,  a  body  started  with  a  velocity  of  40781  + feet  per 
second  from  the  surface  of  the  earth  in  the  direction  of  the  moon 
would  reach  the  centre  of  attraction  and  fall  into  the  moon.  The  ve- 
locity with  which  it  would  arrive  on  the  moon's  surface  is  calculated 
by  the  same  formula,  for  which 

Logarithms. 

£  =  5702400  feet,  the  radius  of  the  moon 6.7560577 

(2=121149600  feet 8.0833220 

£-£  =  1247558400  feet 9.0960611 

£-(2=1132111200  feet 9.0538890 

d- £  =  115447200  feet....  ...  8.0623836 


ASTRONOMY.  305 


In  this  case  J!f=mass  of  the  moon,  and  w  =  mass  of  the  earth. 

The  velocity  with  which  a  body  would  fall  from  the  centre  of  at- 
traction into  the  moon  would  be  only  7468  feet  per  second.  A  body 
thrown  from  the  moon  with  a  velocity  of  7468  feet  per  second  to- 
•\yards  the  earth  would  reach  the  earth  with  a  velocity  of  40781  feet 
per  second. 

The  time  required  for  a  body  to  fall  from  the  centre  of  attraction  to 
the  surface  of  the  earth  is  obtained  by  solving  the  Formula  1,  and 

dd 
placing      F=  —  . 


Find  the  value  of  T  in  this  formula,  which  will  give  the  required 
time  in  seconds. 

26*  U 


APPENDIX. 


DUODENAL   SYSTEM   OF  ARITHMETIC, 
MEASURES,  WEIGHTS  AND  COINS. 

THE  object  of  appending  a  treatise  on  a  new  system  of  arithmetic 
and  metrology  is  to  demonstrate  what  can  be  done  with  that  subject, 
which  demonstration  might  by  that  means  be  conveniently  accessible 
to  the  student  and  to  the  public. 

The  problem  of  an  international  and  complete  system  of  metrology 
has  at  all  times  been  esteemed  an  important  desideratum,  but  no  at- 
tempt has  yet  been  made  to  remove  the  principal  difficulty  which  is 
in  the  way,  and  we  can  expect  no  satisfactory  metrology  until  its 
primary  obstacle  is  removed. 

The  base  ten,  which  is  adopted  in  our  present  arithmetic,  does  not 
admit  of  binary  and  trinary  divisions,  as  required  in  metrology.  This 
is  the  principal  difficulty  in  the  way  of  establishing  a  satisfactory 
system  of  measures,  weights  and  coins. 

The  number  10  is  actually  the  worst  even  number  that  could  have 
been  selected  as  a  base  of  numeration,  for  which  either  8,  12,  or  16 
would  have  been  better. 

The  inconveniences  of  the  decimal  base  in  metrology  are  well 
known,  and  have  been  explained  at  various  times  by  various  writers  ; 
but  the  present  arithmetic  is  so  thoroughly  incorporated  with  civil- 
ization that  it  appears  difficult  to  unlearn  and  get  rid  of  the  same  for 
the  substitution  of  something  better. 

The  American  Pharmaceutical  Association  appointed  a  committee, 
of  which  Alfred*  B.  Taylor  of  Philadelphia  was  chairman,  for  the 
purpose  of  investigating  the  present  condition  of  metrology  with  a 
view  to  its  improvement,  who  gave  the  subject  a  very  careful  and 
deliberate  consideration. 

An  elaborate  report  containing  over  100  octavo  pages  of  fine  print 
was  prepared  and  read  before  the  annual  session  of  the  Association, 
held  in  Boston  September  15,  1859.  This  report  explains  the  incon- 
veniences of  the  decimal  arithmetic  and  of  the  French  metrical  sys- 
tem, illustrated  by  quotations  from  various  authors  of  high  authority. 

In  the  course  of  this  report  Mr.  Taylor  proposed  and  elucidated  an 
Octonal  System  of  arithmetic  and  metrology. 

307 


308  ELEMENTS  OF  MECHANICS. 


Octonal  System. 

The  octonal  system  has  8  to  the  base,  which  admits  of  binary  di- 
vision to  unity  without  fractions.  It  would  be  an  easy  system  to 
learn  and  manage  in  both  arithmetical  and  mental  calculations,  but  it 
requires  a  greater  number  of  figures  than  the  decimal  system  in  ex- 
pressing high  numbers,  and  eight  is  too  small  as  a  base. 

The  octonal  system,  moreover,  does  not  admit  of  trinary  division,  as 
is  required  in  the  circle  and  time. 

Decimal  System. 

The  decimal  arithmetic  is  of  Hindoo  origin,  and  was  imported  into 
Arabia  some  one  thousand  years  ago,  from  which  it  was  spread 
throughout  Europe  and  the  entire  civilized  world. 

The  base  ten  originated  from  the  10  fingers,  which  were  used  for 
counting  before  characters  were  formed  to  denote  numbers. 

The  base  10  admits  of  only  one  binary  division,  which  gives  the 
prime  number  5  without  fraction.  The  trinary  divisions  give  an 
endless  number  of  decimals.  The  decimal  system  is  therefore  not 
well  suited  for  metrology,  in  which  binary  and  trinary  divisions  are 
required. 

It  is  this  defect  of  the  decimal  system  which  has  caused  confusion 
in  metrology  and  discordance  among  nations  respecting  the  adoption 
of  one  common  system  of  measures ;  which  problem  will  never  be 
satisfactorily  solved  as  long  as  decimal  arithmetic  is  maintained. 

By  examining  the  tables  of  measures  and  weights  of  different 
nations  we  find  that  binary  and  trinary  divisions  are  invariably  pre- 
ferred, notwithstanding  that  decimal  arithmetic  must  be  used  in  their 
calculation.  , 

The  French  decimal  metrology  is  perhaps  the  best  that  can  be 
devised  in  connection  with  decimal  arithmetic ;  it  looks  very  inviting 
and  simple  on  paper,  but  what  is  gained  by  the  metrical  system  in 
calculations  is  lost  in  the  shop  and  market. 

The  defects  of  the  metrical  system  are  the  defects  of  our  arithmetic 
itself,  and  as  long  as  decimal  arithmetic  is  maintained  the  French 
system  is  the  best  of  all  that  have  yet  been  devised. 

The  slow  adoption  of  the  metrical  system  by  other  nations  is  sus- 
tained by  good  reasons — namely,  that  it  does  not  constitute  a  com- 
plete, uniform  and  convenient  system  of  metrology.  The  decimal 
system,  as  before  stated,  is  not  applicable  to  the  admeasurement  of  the 
circle,  of  time  and  of  the  compass,  where  binary  divisions  are  indis- 


DECIMAL  SYSTEM.  309 


pensable.  The  circle  requires  both  binary  and  trinary  divisions, 
neither  of  which  can  be  accommodated  by  the  decimal  base. 

When  the  metrical  system  was  first  established  in  France,  it  was 
intended  to  decimate  also  the  circle  and  the 'time,  which  was  soon 
found  to  be  impractical  and  the  idea  abandoned. 

The  French  metrology  is  therefore  not  a  complete  system,  and  it 
has  been  renounced  for  all  measures  in  astronomy,  geography,  navi- 
gation, time,  the  circle  and  the  sphere,  where  it  is  inapplicable. 

The  decimal  system  is  also  inapplicable  in  music,  where  the  binary 
and  trinary  divisions  are  invariably  used. 

Music  represents  the  natural  disposition  of  the  mind  to  arrange  or 
classify  quantities.  The  musical  bar  is  divided  into  halves,  quarters, 
eighths  and  sixteenths ;  and  also  into  thirds,  sixths,  ninths  and  twelfths; 
but  we  never  find  music  divided  into  tenths. 

The  most  natural  or  binary  division  of  music  is  represented  thus : 

Fig.  232. 


A  bar  of  music  divided  by  the  decimal  system  would  appear  thus : 

Fig.  1W.  Fig.  234. 

P    P    P    P    P 

rWrrrfr    orifyouplease 

No  music  could  be  produced  by  either  of  these  last  divisions,  but  a 
mechanical  noise  only  could  be  made  by  it. 

The  lowest  grade  of  man,  and  even  animals,  sing  binary  music. 
Even  an  Australian  magpie  can  be  taught  to  whistle  any  ordinary 
song  as  correctly  as  played  on  a  musical  instrument ;  whereas  a  deci- 
mal division  of  music  could  never  be  learned  and  appreciated  even 
by  the  highest  intelligence. 

Such  is  also  the  comparison  between  binary  and  decimal  arithmetic. 
Decimal  arithmetic  is  a  heavy  burden  upon  the  mind,  and  limits 
mental  calculations  within  a  very  narrow  compass ;  whilst  binary  or 
trinary  arithmetic  would  become  natural  to  the  mind  like  music,  and 
render  mental  calculations  as  easy  as  music  played  by  the  ear. 


310  ELEMENTS  OF  MECHANICS. 


The  Folded  French  Metre. 

The  French  metre  is  difficult  to  fold  into  a  convenient  shape  for  the 
pocket.  The  ten-folded  metre  with  lap-joints  is  a  very  convenient 
form  for  approximate  measurements,  but  cannot  be  relied  upon  for 
correctness,  because  the  numerous  lap-joints  cannot  be  made  perma- 
nently accurate,  and  moreover  the  lap-joints  do  not  form  a  straight 
but  a  broken  line.  The  metre  folded  into  five  parts  with  lap-joints 
is  an  odd  affair. 

The  two-folded  metre  of  five  decimetres  in  each  part,  of  about  20 
inches  long,  is  too  large  for  the  pocket. 

The  four-folded  metre  makes  two  and  a  half  decimeti'es  in  each 
part  of  about  10  inches  long,  which  will  answer  for  the  pocket ;  and 
is  perhaps  the  best  form  of  the  French  metre  when  made  with  regular 
hinges  like  the  English  four-folded  rule,  but  it  is  still  a  broken 
measure. 

An  international  association  for  obtaining  a  uniform  decimal  system 
of  weights,  measures  and  coins  has  been  in  existence  for  over  thirty 
years,  and  has  yet  accomplished  very  little.  The  object  of  this  asso- 
ciation is  wholly  for  the  introduction  of  the  French  metrical  system, 
which  has  met  with  the  most  natural  and  reasonable  objections — 
namely,  that  it  is  not  a  complete  system,  and  that  it  is  inconvenient 
in  the  shop  and  in  the  market ;  but  the  strong  influence  of  this  asso- 
ciation has  induced  many  governments  to  force  that  system  upon 
their  people. 

In  practice,  we  want  our  units  divided  into  the  simplest  and  most 
natural  fractions — namely,  halves,  thirds,  quarters,  sixths,  eighths,  etc. — 
which  cannot  be  done  by  the  metrical  system,  or  decimal  arithmetic 
without  long  tails  of  figures  commencing  with  0. 

For  instance,  the  simple  fraction  %  expressed  by  decimals  is 

0.33333 without  end,  and  will  never  be  correct,  and  requires  a 

good  education  to  understand  the  true  meaning  of  it.  The  good 
scholar  manages  the  decimal  fractions  as  easily  as  a  musician  plays  on 
his  hand-organ,  but  the  fraction  0.33333  is  not  so  easily  understood 
by  the  majority  of  the  people,  who  will  naturally  ask  what  it  means. 
In  the  answer  it  is  necessary  to  explain  that  the  unit  is  divided  into 
100000  parts,  and  33333  of  those  parts  is  nearly  £  of  the  whole.  The 
people  will  then  surely  reply  that  they  are  not  willing  to  cut  their 
things  up  into  100000  parts  and  lose  a  portion  by  the  division  in 
order  to  get  it  into  three. 


DUODENAL  SYSTEM.  311 


Duodenal  System. 

Charles  XII.  of  Sweden  proposed  to  introduce  a  duodenal  system 
of  arithmetic  and  metrology.  The  king  complained  of  ten  as  a  base, 
and  said,  "  It  can  be  divided  only  once  by  2,  and  then  stops."  The 
number  12  can  be  divided  by  2,  3,  4  and  6  without  leaving  fractions ; 
and  divided  by  8  gives  f ,  by  9  gives  f ,  and  by  10  gives  f ,  all  con- 
venient fractions  for  calculation. 

The  number  12  has  always  been  a  favorite  base  in  metrology. 

The  old  French  foot  was  divided  into  12  inches,  the  inch  into  12 
lines,  and  the  line  into  12  points.  The  dozen  is  a  well-known  base 
adopted  all  over  the  world ;  12  dozens  is  a  gross,  and  12  gross  is  a 
great-gross.  We  have  12  months  in  a  year,  12  hours  in  a  day,  12 
signs  in  the  zodiac,  12  musical  notes  in  an  octave.  The  old  Roman 
metrology  was  based  on  12,  like  the  English  foot  and  the  Troy  pound. 

A  writer  in  the  Edinburgh  Review  (Jan.,  1807,  vol.  9,  page  376} 
regrets  that  the  philosophers  of  France,  when  engaged  in  making  so 
radical  a  change  in  the  measures  and  standards  of  the  nation,  did  not 
attempt  a  reform  in  the  popular  arithmetic.  He,  being  in  favor  of  a 
duodenal  system,  says,  "  The  property  of  the  number  12  which  re- 
commends it  so  strongly  for  the  purpose  we  are  now  considering  is  its 
divisibility  into  so  many  more  aliquot  parts  than  ten,  or  any  other 
number  that  is  not  much  greater  than  itself.  Twelve  is  divisible  by 
2,  3,  4  and  6;  and  this  circumstance  fits  it  so  well  for  the  purpose  of 
arithmetical  computation  that  it  has  been  resorted  to  in  all  times  as 
the  most  convenient  number  into  \vhich.  any  unit  either  of  weight  or 
measure  could  be  divided.  The  divisions  of  the  Roman  as,  the  libra, 
the  jugerum,  and  the  modern  foot,  are  all  proofs  of  what  is  here  as- 
serted ;  and  this  advantage,  which  was  perceived  in  rude  and  early 
times,  would  have  been  found  of  great  value  in  the  most  improved 
eras  of  mathematical  science.  .  .  .  We  regret  therefore  that  the  ex- 
periment of  this  new  arithmetic  was  not  attempted.  Another  op- 
portunity of  trying  it  is  not  likely  to  occur  soon. 

"  In  the  ordinary  course  of  human  affairs  such  improvements  are  not 
thought  of,  and  the  moment  may  never  again  present  itself  when  the 
wisdom  of  a  nation  shall  come  up  to  the  level  of  this  species  of  reform." 

If  man  had  been  created  with  six  fingers  on  each  hand,  we  would 
have  had  in  arithmetic  a  duodenal  instead  of  the  present  cumbrous 
decimal  system. 

A  uniform  duodenal  system  of  metrology,  even  with  decimal  arith- 
metic, would  be  much  better  in  the  shop  and  market  than  the  French 
metrical 


312 


ELEMENTS  OF  MECHANICS. 


A  duodenal  system  would  be  equally  applicable  in  all  branches  of 
metrology,  and  it  would  include  those  which  are  excluded  by  the 
metrical  system — namely,  astronomy,  geography,  navigation,  time  and 
the  circle. 

The  duodenal  system  would  require  two  new  characters  to  repre- 
sent 10  and  11,  so  as  to  place  10  at  12.  This  change  in  the  figures 
would  appear  strange  at  the  first  glance,  but  a  little  reflection,  with 
due  consideration,  would  soon  lead  to  the  satisfaction  that  these  two 
new  figures  simplify  the  arithmetic  and  render  it  much  easier  for 
mental  calculation  than  decimal  arithmetic. 


Senidenal   System. 

The  senidenal  system  has  16  to  the  base.  A  full  elucidation  of 
this  system  has  been  worked  out  by  the  author  and  was  published  in 
the  year  1862  by  J.  B.  Lippincott  &  Co.,  Philadelphia.  It  is  called 
the  tonal  system. 

The  advantage  of  16  as  a  base  for  arithmetic  is  that  of  its  binary 
division  to  infinity.  It  is  really  the  best  system  that  could  be  devised 
for  metrology  and  mental  calculations. 

The  disadvantage  of  16  as  a  base  is  that  it  requires  six  new  figures 
to  complete  the  base,  which  would  be  difficult  to  introduce,  and  also 
that  it  does  not  admit  of  trinary  divisions,  as  is  required  in  the  circle 
and  time,  but  it  is  under  all  circumstances  far  superior  to  the  decimal 
system. 

The  difficulties  with  the  decimal  system  are  fully  explained  in  the 
elucidation  of  the  tonal  system. 

Scale  of  Four  Arithmetical  Systems. 


Systems. 

Base. 

100 

1000 

10,000 

100,000 

1,000,000 

Octonary  

8 
10 

64 

100 

512 

1,000 

4,096 
10,000 

32,768 
100,000 

260,744 
1,000,000 

Duodenary  
Senidenary  

12 
16 

144 
256 

1,728 
4,096 

20,736 
65,536 

238,832 
16,777,216 

2,865,984 
268,435,456 

The  names  of  the  systems  are  Hindoo. 

The  octonary  system  requires  the  greatest  number  of  places  for  ex- 
pressing high  numbers,  for  instance  1,000,000  octonal  means  only 
260,744  of  the  decimal  system. 

The  senidenary  or  tonal  system  uses  less  places;  for  instance, 
1,000,000  senidenal  means  268,435,456  of  decimal  numbers. 


NOMENCLATURE.  313 


DUODENAL  ARITHMETIC  AND    METROLOGY. 

The  base  in  the  duodenal  system  is  12,  instead  of  10  in  the  decimal 
system. 

The  Arabic  system  of  notation  is  composed  of  ten  simple  digits,  or 
characters — namely,  0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  and  the  base  10. 

These  same  characters  can  be  used  in  the  duodenal  system  by  add- 
ing two  numbers  to  complete  the  base — namely,  11  and  12 ;  then  all 
the  units  of  weights  and  measures  should  be  divided  and  multiplied 
by  12,  but  in  order  to  render  the  system  simple  for  calculation,  it  will 
be  necessary  to  substitute  new  characters  for  the  numbers  10,  11  and 
12 — namely, 

Decimal  system,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10,  11,  12 ; 

Duodenary  system,          1,  2,  3, 4,  5,  6,  7,  8,  9, 9?,  5,  10, 

in  which  10  denotes  the  base  12,  $  stands  for  10,  and  °S  stands  for  11. 

The  Italic  figures  mean  decimal  numbers,  and  the  Roman  figures 
mean  duodenal  numbers. 

In  order  to  distinguish  the  two  systems  from  one  another,  it  will 
be  necessary  to  give  new  names  to  the  duodenary  figures. 

A  duodenary  system  of  arithmetic  cannot  be  adopted  by  only  one 
nation,  but  the  whole  civilized  world  ought  to  agree  upon  such  a 
scheme.  Different  nations  have  different  languages  and  names  for 
the  decimal  figures  and  numbers ;  but  in  the  adoption  of  a  duodenary 
system  of  arithmetic,  one  common  nomenclature  might  be  agreed  upon. 

The  new  figures  and  nomenclature  appear  to  be  the  greatest  objec- 
tion to  the  introduction  of  the  duodenal  system  of  arithmetic  and 
metrology. 

There  is  no  difficulty  in  convincing  the  public  of  the  utility  of  the 
duodenal  system,  and  with  that  impression,  a  pride  will  be  taken  in 
using  the  new  nomenclature,  which  could  be  taught  in  every  school ; 
and  each  individual  would  attempt  to  follow  up  the  time  of  education. 

The  following  table  contains  the  names  of  the  figures  and  numbers 
up  to  twelve  in  different  languages : 


314 


ELEMENTS  OF  MECHANICS. 


Nomenclature  of  Numbers  in  Different  Languages. 

No. 

English. 

French. 

German. 

Swedish. 

Spanish. 

Latin. 

Greek. 

0 

Naught. 

Z6ro. 

NulL 

Noll. 

Cero. 

Nihil. 

Zero. 

i 

One. 

Un. 

Eins. 

En. 

Uno. 

TJnus. 

Eis,  En. 

2 

Two. 

Deux. 

Zwei. 

Tvi 

Dos. 

Duo. 

Duo. 

Three. 

Trois. 

Drel. 

Tre. 

Tres. 

Tres. 

Treis. 

Four. 

Quatre. 

Vier. 

Fyra. 

Cuatro. 

Quatuor. 

Tessares. 

Five. 

Cinq. 

Funf. 

Fern. 

Cinco. 

Quinque. 

Pente. 

Six. 

Six. 

Sechs. 

Sex. 

Seis. 

Sex. 

Hex. 

Seven. 

Sept. 

Siben. 

Sju. 

Siete. 

Septem. 

Kept  a. 

Eight. 

Huit. 

Acht. 

Atta. 

Ocho. 

Octo. 

Okto. 

Nine. 

Neuf. 

Neun. 

Nio. 

Nueve. 

Novem. 

Ennea. 

10 

Ten. 

Dix. 

Zehn. 

Tio. 

Diez. 

Decem. 

Deka. 

11 

Eleven. 

dn/o. 

Elf. 

Elva. 

Once. 

Undecem. 

Endeka. 

12 

Twelve. 

Douze. 

Zwolf. 

Tolf. 

Doce. 

Duodecem. 

Dodeka.   ; 

No. 

Russian. 

Finnish. 

Welsh. 

Gaelic. 

Gothic. 

Turkish. 

Hebrew. 

Nill. 

Nolli. 

Zuffer. 

Ay.in. 

Odna. 

YksL 

Una. 

Unah. 

Ains. 

Bier. 

Aleph. 

Dva. 

Kaksi. 

Dou. 

Gha. 

Tvai. 

Icki. 

Beth. 

Tri. 

Koleme. 

Tri. 

Tree. 

Threis. 

Utch. 

Gimel. 

Tcheteri. 

Nelja. 

Pedvar. 

Cheir. 

Fidvor. 

Duert. 

Daleth. 

Fiat. 

Viisi. 

Pump. 

Coag. 

Finf. 

Bach. 

He. 

Shest 

Kunsi. 

Chewech. 

Seach. 

Saihs. 

Altoe. 

Vau. 

7 

Sem. 

Seitseman. 

Saith. 

Sheach. 

Sibum. 

Yedi. 

Zam. 

8 

Vosem. 

Kahdeksan. 

Wyth. 

Oacht. 

Ahtan. 

Seckiz. 

Bheth. 

9 

Devi  at. 

Yhdeksan. 

Nan. 

Nuegh. 

Niun. 

Dokus. 

Teth. 

10 

Desiat. 

Kymmenan. 

Deg. 

Doach. 

Taihun. 

On. 

Yod. 

11 

Odenatset. 

Yksitoista. 

Undeg. 

Undech. 

Ainstaihun. 

On  bier. 

Yodaleph. 

12 

Dvenatset. 

Kaksitoista. 

Doudeg. 

Ghadech. 

Tvaitaihun. 

Onicki. 

Yodbeth. 

No. 

Arabian. 

Persian. 

Hindoo. 

Chinese. 

Japanese. 

Sanscrit. 

Duodenal 

0 

Siforon. 

Bow. 

Ley. 

Zero,  0 

1 

Ahed. 

Yika. 

Ek  Ache. 

Yat. 

Itchi. 

Aika. 

An,     1 

2 

Ishnan. 

Du. 

Do. 

Ge. 

Ni. 

Dwan. 

Do,     2 

1 

Saylaset. 

Seh. 

len. 

Sam. 

San. 

Tri. 

Tre,    8 

4  !  Erbayet. 

Chehaur. 

ChaT. 

Tze. 

TchL 

Chatur. 

For,   4 

:>  1  Jemset 

Pendj. 

Panoh. 

Ngnn. 

Go. 

Pancha. 

Pat,  5 

6  |  Sittet. 

Shesh. 

Chha. 

Luck. 

Lock. 

Shash. 

Sex,   6 

't  \  Saybet. 

Helft. 

Sath. 

Tchut. 

Sytchi. 

Saptan. 

Ben,   7 

S  i  Saymaniet. 

Hesht, 

Ath. 

Pbat. 

Hatchi. 

Ashta. 

Ott,    8     ! 

9  |  Tiset. 

Nuh. 

Nau. 

Geo. 

Ku. 

Navan. 

Net,    9    ,' 

10 

Eshret. 

Deh. 

Das. 

Shop. 

Dgiu. 

Dashan. 

Dis,  ? 

11 

Ahedeshere. 

Yikadeh. 

Gyarah. 

Shopyat. 

Dgiuitchi. 

Aikadashan. 

Elr,  "S 

12 

Ishnaueshere. 

Dudeh. 

Barah. 

Shopgaa. 

Dgiuni. 

D\vandashan. 

Ton,10 

| 

NOMENCLATURE.  315 


COMMENTS  ON  NOMENCLATURE. 

On  account  of  the  different  pronunciations  of  letters  and  words  in 
different  languages,  the  true  sound  of  a  name  cannot  be  conceived 
without  a  knowledge  of  the  language  in  which  it  is  written. 

The  Japanese  sound  for  9  is  written  leu  in  the  table,  but  for  the 
English  pronunciation  it  should  be  written  koo. 

There  are  some  letters  of  the  alphabet  which  have  nearly  the  same 
sound  in  all  languages,  and  only  such  letters  should  be  used  in  the 
coining  of  names  for  the  figures  and  numbers  in  the  duodenary  system. 

The  letters  th,  w,  o,  ur,  ght  in  the  English  language,  and  also  the 
letter  C,  which  has  two  sounds  in  almost  all  languages,  should  not  be 
used  for  the  new  names. 

The  names  given  to  the  duodenary  figures  in  the  last  column  are 
clear  and  distinct  sounds,  -which  would  be  well  understood  and  pro- 
nounced alike  in  all  languages. 

It  would  be  useless  to  attempt  to  introduce  the  names  of  the  figures 
and  numbers  in  either  of  the  languages  above  given  as  a  universal 
nomenclature,  for  not  only  that  they  are  net  suited  for  more  than  the 
language  in  which  they  are  writtea,  but  prejudices  would  be  against 
them.  The  introduction  of  the  French  metrical  system  has  been 
greatly  retarded  by  reason  merely  of  its  cumbersome  nomenclature. 

The  best  work  on  the  etymology  of  -numbers  known  to  the  author 
is  that  of  Professor  S.  Zehetmayr,  published  in  Leipsic,  1854.  In 
the  establishment  of  a  new  and  universal  nomenclature  of  numbers 
we  ought  to  select  clear  and  distinct  sounds,  which  can  be  understood 
and  pronounced  alike  in  all  languages,  without  regard  to  the  ety- 
mology of  numbers. 

The  Arabic  notation  of  numbers  is  yet  used  only  by  about  one-third 
of  the  population  of  the  earth,  and  the  other  two-thirds  use  different 
kinds  of  irregular  characters  or  hieroglyphics,  which  combinations 
are  unfit  for  arithmetical  calculations. 

The  Roman  notation  was  used  in  England  up  to  the  beginning  of 
the  seventeenth  century,  when  the  Arabic  notation  was  gradually 
gaining  ground  against  very  strong  opposition ;  and  at  last  caused 
the  burning  of  the  houses  of  Parliament.  The  Arabic  notation  was 
introduced  into  Germany  in  the  twelfth  century,  and  into  Italy  in  the 
eleventh  century. 


316 


ELEMENTS  OF  MECHANICS. 


Comparison  of  Numbers  in  the  Duodenary  and  Decimal 
Systems,  with  the  Corresponding  New  Names. 


New. 

Names. 

Old. 

New. 

Names. 

Old. 

o 

Zero  

0 

37 

Tretoben  

43 

1 

An. 

1 

38 

Tretonott      ..     . 

44 

2 

Do 

<J> 

39 

Tret  on  ev 

45 

3 

Tre 

3 

3T 

Tretondis 

46 

4 

For  

4 

35 

Tretonelv  

47 

5 

Pat  

5 

40 

Forton  . 

48 

6 

Sex 

6 

41 

Fortonan 

49 

7 

Ben  

7 

42 

Fortondo  

50 

8 

Ott  

8 

43 

Fortontre  

51 

9 

Nev... 

9 

44 

Fortonfor 

52 

$ 

Dis 

10 

45 

Fort  on  pat 

53 

8 

Elv  

11 

46 

Fortonsex  

54 

10 

Ton  

12 

47 

Fortoben  

55 

11 

Tonan  . 

13 

48 

Fortonott 

56 

12 

Tondo 

14 

49 

Fortonev 

57 

13 

Tontre  

15 

4$ 

Fortondis  

58 

14 

Tonfor  .. 

16 

45 

Fortonelv  .... 

59 

15 

Tonpat 

17 

50 

Paton  . 

60 

16 

Tonsex 

18 

51 

Patonan 

61 

17 

Toben  

19 

52 

Patondo  

62 

18 

Tonott 

20 

53 

Patontre  

63 

19 

Tonev 

21 

54 

Patonfor 

64 

1$ 

Tondis 

22 

55 

Patonpat 

65 

15 

Tonelv  

23 

56 

Patonsex  ] 

66 

20 

Doton  .  . 

24 

57 

Patoben  

67 

21 

Dotonan  

25 

58 

Patonott  

68 

22 

Dotondo 

26 

59 

Patonev 

69 

23 

Dotontre  

27 

5$ 

Patondis  

70 

24 

Dotonfor 

28 

55 

Paton  elv      

71 

25 

Doton  pat 

29 

60 

Sexton  . 

72 

26 

Dotonsex 

30 

61 

Sextonan 

73 

27 

Dotoben    

31 

62 

Sextondo  

74 

28 

Detonott 

32 

63 

Sextontre  

75 

29 

Detonev 

33 

64 

Sextonfor 

76 

2$ 

Dotondis 

34 

65 

Sextonpat 

77 

25 

Dotonelv 

So 

66 

Sextoneex 

78 

30 

Treton  .... 

36 

67 

Sextoben  

79 

31 

Tretonan 

37 

68 

Sextonott  

80 

32 

Tretondo 

38 

69 

Sextonev 

81 

33 

Tretontre 

39 

6$ 

Sextondis 

82 

34 

Tretonfor  

40 

65 

Sextonelv  

83 

35 

Tretonpat 

41 

70 

Benton  

84 

36 

Tretonsex 

42 

71 

Bentonan 

85 

NOMENCLATURE    OF    NUMBERS. 


317 


New. 

Names. 

Old. 

New. 

Names. 

Old. 

72 

Bentondo  

86 

51 

Elvtonan  

133 

73 

Bentontre  

87 

ft2 

Elvtando  

134 

74 

Bentonfor  

88 

ft3 

Elvtontre  

135 

75 

Bentonpat  

89 

ft4 

El  vtonfor  

136 

76 

Bentonsex  .  .  . 

90 

ft5 

Elvtonpat    . 

137 

77 
78 

Bentoben  
Bentonott 

91 

92 

"86 

ft7 

Elvtoneex  
Elvtoben 

138 
139 

79 

Bentonev  

93 

#8 

Elvtonott  

140 

7f 

Bentondis  

94 

ff9 

Elvtonev  

141 

7ft 

Bentonelv  

95 

3f 

Elvtondis 

142 

80 

Otton  .       .   . 

96 

M 

Elvtonelv 

143 

81 

Ottonan    > 

97 

100 

San. 

H4 

82 

Ottondo 

98 

148 

San-fortonott 

200 

83 

Ottontre  

99 

200 

Dosan  

288 

84 

Ottonfor  

100 

210 

Dosan-ton  

300 

85 

Ottonpat  

101 

300 

Tresan  

432 

86 

Ottonsex 

102 

358 

Tresan-patonott 

500 

87 

Ottoben 

103 

400 

Forsan 

576 

88 

Ottonott 

104 

420 

Forsan-doton 

600 

89 

Ottonev  

105 

500 

Patsan  

720 

8? 

Ottondis  

106 

568 

Patsan-sextonott 

800 

8ft 

Ottonelv 

107 

600 

Sexan    

864 

90 

Nevton 

108 

630 

Sexan-treton 

900 

91 

Nevtonan 

109 

700 

Bensan 

1008 

92 

Nevtondo  

110 

800 

Ottsan  

1152 

93 

Nevtoutre  

111 

900 

Nevsan  

1296 

94 

Nevtonfor  

112 

fOO 

Dissan  

1440 

95 

Nevtonpat    . 

113 

ftOO 

Elvsan  

1584 

96 

Nevtonsex 

114 

1000 

Tos 

1728 

97 

Nevtoben 

115 

1100 

Tossan 

1872 

98 

Ne  vtonott  

116 

1200 

Tosdosam  

2016 

99 

Nevtonev  

117 

1300 

Tostresan  

2160 

9? 

Nevtondis 

118 

1400 

Tosforsan  

2304 

9ft 

Nevtonelv 

119 

1500 

Tospatsan     

2448 

$0 

Diston 

120 

1600 

Tossexan 

2592 

K 

Distonan 

121 

1700 

Tosbensan 

2736 

f2 

Distondo  

122 

1800 

Tosottsan  

2880 

9?3 

Distontre  

123 

1900 

Tosnevsan  

3024 

$4 

Distonfor  . 

124 

IfOO 

Tosdissan  

3168 

$5 

Distonpat 

125 

IftOO 

Toselvsan     ... 

3312 

$6 

Distonsex 

126 

2000 

Dotos 

3456 

$7 

Distoben 

127 

4000 

Fortos 

6912 

$8 

Distonott  

128 

6000 

Sextos  

10368 

$9 

Distonev    

129 

8000 

Ottos  

13724 

$f 

Distondis 

130 

?000 

Distos  

17180 

<Pft 

Dirfonelv 

131 

10000 

Dill  

20736 

ftO 

Elvton  

132 

318 


ELEMENTS  OF  MECHANICS. 


FRACTIONS. 

Duodenary  System.  Decimal   System. 


|  =  0.9 
|  =--0.16 
|  =  0.46 
f  =  0.76 

|  =  0.4 
f  =  0.8 
i  =  0-2 

^  =  0.09 
ft  =  0.23. 
i^  =  0.53 
£.  =  0.83 
^  =  0.06 
^  =  0.36 

£  =  0.56 

3&  =  0.046 

A  =  0.976 

|  ="0.14 

|  =  0.24 

|  =  0.68 
|  =  0.?S 

£  =  0.416 
£|  =  0.646 
^-0.023 

^  =  0.209 
ff- 0.739 


=  0.75 


1  =  0.375 
§  =  0.625 
1  =  0.875 


1  =  0.6666.  .  . 

%  =  0.16666.  . 

§  =  0.83333.  . 
&  =  0.0625 

|  =  0.1875 
Js  =  0.4375 
Q  =  0.6875 
£f  =  0.041666. . 
fa  =0.29 16666. 
1^=0.458333.. 
jy  =  0.03-125 
^  =  0.21875 

$  =  0.11111111 

$  =  0.22222 

§  =  0.44444-  • 
§  =  0.55555.  . 
\  =  0.88888.  . 
%%  =  0.34375 
£f  =  0.53125 
fo  =  0.015625 
#1  =  0.171875 
11  =  0.609375 


The  abovre  table  of  fractions  shows  the  simplicity  of  the  duodenary 
system,  which  requires  few  figures  where  the  old  system  requires  a 


DUODENAL  ARITHMETIC. 


319 


great  number  of  decimals.  For  3ds,  6ths,  9ths,  12ths  and  24ths  the 
duodenary  system  finishes  the  fraction  with  one  or  two  places  where 
the  number  of  decimals  is  endless. 


Addition  Table. 


1 

2 

3 

4 

5 

6 

7 

8 

9 

$ 

8 

10 

2 

4 

5 

6 

7 

8 

9 

$ 

V 

10 

11 

12 

3 

5 

6 

7 

8 

9 

$ 

8 

10 

11 

12 

13 

4 

6 

i 

8 

9 

f 

S 

10 

11 

12 

13 

14 

5 

7 

8 

9 

¥ 

7$ 

10 

11 

12 

13 

14 

15 

G 

8 

9 

$ 

8 

10 

11 

12 

13 

14 

15 

16 

7 

9 

$ 

8 

10 

11 

12 

13 

14 

15 

16 

17 

8 

« 

8 

10 

11 

12 

13 

14 

15 

16 

17 

18 

9 

V 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

f 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

IT 

8 

11 

12 

13 

14 

15 

16 

17 

18 

19 

IT 

US 

10 

12 

13 

14 

15 

16 

17 

18 

19 

1? 

18 

20 

Multiplication  Table. 

2 

3 

4 

5 

6 

7 

8 

9 

f 

8 

10 

2 

4 

8 

I 

10 

12 

14 

16 

18 

If 

20 

3 

6 

9 

10 

13 

16 

19 

20 

23 

26 

29 

30 

4 

8 

10 

14 

18 

20 

24 

28 

30 

34 

38 

40 

5 

IP 

13 

18 

21 

26 

2ff 

34 

39 

42 

47 

50 

6 

10 

16 

20 

26 

20 

36 

40 

46 

50 

56 

60 

7 

12 

19 

24 

2ff 

36 

41 

48 

53 

5$ 

65 

70 

8 

14 

20 

28 

34 

40 

48 

54 

60 

68 

74 

80 

9 

16 

23 

30 

39 

46 

53 

60 

69 

76 

83 

90 

5 

18 

26 

34 

42 

50 

5$ 

68 

76 

84 

92 

$0 

ff 

1$ 

29 

38 

47 

56 

65 

74 

83 

92 

fl 

TO 

10 

20 

30 

40 

50 

60 

70 

80 

90 

$0 

TO 

100 

, 

The  duodenal  multiplication  table  of  the  single  figures  is  44  per 
cent,  more  extensive  than  that  of  the  decimal,  but  the  binary  and 
trinary  properties  makes  it  much  easier  to  learn  and  to  remember. 


320 


ELEMENTS  OF  MECHANICS. 


3854553 
6108555 
9951598 


Examples  in  Addition. 

85093 


535 

3$ 
98555 


0.03 
53.06 

459.51 

6745.60 

39506.00 

44536.45 


745856 
314364 
437542 


8694 
24 

25314 
5168 
75994 


Examples  in  Subtraction. 

3543.51  0.54856 

1595.05  0.00355 


1565.53 
Examples  in  Multiplication. 

$4563 
635 


0.54458 


956689 
272569 
525916 
55536759 

Examples  in  Division. 


365.3545 
0.056 
19575226 
32442507 
34.1954296 


42)136505(3854 
106 
305 
294 


45.5)35057.635(946.38 
3823 
1527 

1778 


370 
358 
145 
148 
002 


2556 
1603 
1289 
3364 
3335 
36 


On  account  of  the  binary  and  trinary  properties  of  the  duodenary 
system,  these  arithmetical  operations  are  much  easier  to  the  mind  than 
those  with  decimal  arithmetic.  The  only  difficulty  about  it  is  to  un- 
learn the  decimal  system. 

The  duodenary  system  has  all  the  advantages  and  none  of  the  dis- 
advantages of  the  decimal  system  ;  it  is  also  better  adapted  to  mental 
calculations,  which  are  very  difficult  with  our  present  arithmetic. 


DUODENAL  METROLOGY.  321 


METROLOGY. 

The  utility  of  a  duodenary  system,  of  arithmetic  consists  in  its  com- 
bination with  a  similar  system  of  metrology — namely,  that  all  units 
of  measure  should  be  divided  and  multiplied  by  the  same  base,  twelve. 

Units  of  measure  are  required  for  the  following  fifteen  quantities. 


Length. 
Surface. 
Volume. 


Weight. 

Mass. 

Money. 


Heat. 
Light. 
Electricity. 


Force. 

Velocity. 

Time. 


Power. 
Space. 
Work. 


Measurement  of  Length. 

Assume  the  mean  circumference  of  the  earth  to  be  the  primary 
unit  of  length,  and  divide  it  by  twelve  repeatedly  until  the  divisions 
are  reduced  to  a  length  which  would  be  a  convenient  unit  to  handle 
in  the  shop  and  in  the  market. 

The  mean  circumference  of  the  earth  is  about  24851.64  miles, 
which,  multiplied  by  5280,  will  be 

Duodenal. 

0  131216659.2  feet 1  circum. 

1  10934721.6  feet 1  hour. 

2  911226.8  feet ,....     1  grad. 

3  75985.56  feet 1  minute. 

4  6327.96  feet 1  mile. 

5  527.33  feet 1  cable. 

6  43.944  feet 1  chain. 

7  3.772  feet 1  metre. 

The  required  unit  of  length  43.944  inches 1  metre. 

The  length  of  the  circumference  of  the  earth,  divided  by  the  seventh 
power  of  12,  gives  a  length  of  4$-944  inches,  which  is  assumed  as  a 
unit  for  all  measurements  of  length,  and  which  we  will  call  a  metre. 

Twelve  duodenal  metres  is  a  length  of  4$ -944  feet,  which  is  a  convenr 
ient  measure  in  the  field  or  in  surveys,  and  which  we  will  call  a  chain. 

Twelve  duodenal  chains  is  a  length  of  527.33  feet,  which  we  will 
call  one  cable. 

Twelve  duodenal  cables  is  a  length  of  6327.96  feet,  which  we  will 
call  one  mile.  The  duodenal  mile  will  be  about  300  feet  longer  than 
our  present  knot  or  sea-mile. 

Twelve  duodenal  miles      =  1  minute,  ^ 

Twelve  duodenal  minutes  =  1  grad,        I  Qn  ^  earth,s         t  ^^ 
Twelve  duodenal  grads     =  1  hour, 
Twelve  duodenal  hours     =  1  circum,  ' 
V 


322 


ELEMENTS  OF  MECHANICS. 


The  duodenal  metre  to  be  divided  into  twelve  equal  parts  of  3.772 
inches  each,  and  called  metons.  The  meton  into  twelve  equal  parts 
of  0.31433  of  an  inch  each,  called  mesans.  The  mesan  into  twelve 
equal  parts  of  0.0262  of  an  inch  each,  called  metos. 

Fig.  235  shows  the  full  size  of  a  meton  with  its  divisions. 


Fig.  235. 


Li,.i 


f     ?    8    *    5    f    f     »    ir 

.....  I..I..T..I..L.I..I..I..I..T..I..U.T  .....  i..i..r.i..i  1  1  1  1  1  1  1  1  1  T  1  1  1  1  1  1  1 


10 


ii 


The  first  6  mesans  are  divided  into  metos,  and  the  last  into 
quarters  of  mesans.  The  ordinary  shop-metre  need  not  be  divided 
finer  than  into  quarters  of  mesans,  for  in  so  small  divisions  the 
metos  can  easily  be  approximated. 

The  metons  and  mesans  would  be  the  most  convenient  for  express- 
ing short  measures  in  the  mechanic  arts. 


Fig.  236. 


Fig.  240. 


Fig.  236  represents  a  twelve-folded  duodenal  metre  with  lap-joints, 
like  the  ten-folded  French  metre ;  each  part  is  one  meton  of  3.772 
inches. 

Fig.  237  represents  a  six-folded  duodenal  metre  with  lap-joints,  of 
7.544  inches  in  each.  This  form  could  be  made  with  regular  hinges 
like  the  English  rule. 

Fig.  238  represents  a  four-folded  duodenal  metre,  with  3  metons  in 
each  part  of  11.316  inches.  This  would  be  the  most  convenient  form 
for  the  shop  when  folded  with  regular  hinges  like  the  English  four- 
folded  rule 


DUODENAL  METROLOGY. 


323 


Fig.  239  represents  a  three-folded  metre,  with  four  metons  in  each 
part  of  15.088  inches. 

Fig.  240  represents  a  two-folded  metre,  with  six  metons  in  each  part 
of  about  22  inches. 

We  see  here  that  the  duodenal  metre  can  be  folded  into  five  dif- 
ferent forms,  with  even  measures  in  each  part. 

The  longest  unit  of  measure  is  the  circumference  of  the  earth, 
which  ought  to  be  termed  a  circum.  The  circum  should  be  used  in 
expressing  astronomical  distances. 

The  duodenal  grad  is  100  duodenal  miles,  or  0.01  of  the  earth's 
great  circle,  which  would  be  a  proper  measure  for  expressing  long 
distances  on  the  earth's  surface ;  asd  which  would  convey  a  correct 
idea  of  the  real  magnitude  of  such  distances  compared  with  the  great 
circle.  ,  . 

The  mile  would  be  the  common  road  measure  and  for  traveling 
distances  on  land  and  sea. 


Duodenal  Measures  of  Length. 


Circum. 

Grad. 

Mile. 

can 

'Chain. 

"Metre. 

Meton. 

Mesan. 

Metos. 

1 

100 

10000 

100000 

1000000 

0.01 

1 

100 

1000 

10000     100000 

1000000 

0.0001 

0.01 

1 

10 

1600 

10000 

100000 

1000000 

0.0001 

0.1 

1 

10 

1000 

10000 

100000 

1000000 

0.000001 

0.01 

0.1 

a 

10 

1000 

10000 

100000 

0.001 

0.01 

0.1 

1 

10 

1000 

10000 

0.0001 

0,001 

0.01 

H».l 

1 

10 

1000 

0.00001 

0.0001 

0.001 

0.01 

0.1 

1 

10 

o.ooooi  !  o.oooi 

0.001 

0.01 

0.1 

1 

Division  of  tlie  Circle. 

The  circle  to  be  divided  into  100  equal  parts  (144  decimal). 


Duodenal  System. 
One  circle        =  100  grads. 
One  grad         =  100  lento. 
One  lent          =  100  ponts. 
One  pont          = 
One  quadrant  =    30  grads. 


Old  System. 


2  degrees  30  minutes. 
1  minute  2.5  seconds. 
0.43418  of  a  second. 
90  degrees. 


One  duodenal  mile  on  the  earth's  surface  corresponds  with  an  angle 
of  one  lent. 


324  ELEMENTS  OF  MECHANICS. 

One  duodenal  chain  on  the  earth's  surface  corresponds  with  an 
angle  of  one  pont. 

The  latitude  and  longitude  to  be  divided  as  the  circle. 

The  angular  measures  correspond  with  the  linear  measures  on  the 
earth's  surface.  The  terms  minute  and  second  are  omitted  in  the  di- 
vision of  the  circle,  so  as  not  to  confound  angles  with  time. 

The  circle  can  thus  be  divided  into  2,  3,  4,  6,  8,  9, 12  or  16  parts, 
without  leaving  fractions  of  a  degree  or  grad. 

The  quadrant  of  the  circle,  containing  30  grads  (36),  can  be  divided 
into  2,  3,  4,  6,  9  or  12  parts  without  leaving  fractions  of  a  grad. 
These  advantages  with  the  duodenal  division  of  the  circle  are  of 
great  importance  in  geometry,  geography,  trigonometry,  astronomy 
and  in  navigation. 

Either  of  the  divisions  corresponds  with  an  even  linear  measure  on 
•the  earth's  surface. 

Duodenal  Division  of  Time. 

The  division  of  time  should  conform  to  that  of  the  circle. 
The  time  from  noon  to  noon,  including  one  night  and  day,  to  be 
divided  into  twelve  equal  parts,  called  hours. 

Duodenal  System.  Old  System. 

One  day       =    10  hours.  24  hours. 

One  hour     =   10  grads.  2  hours. 

One  grad     =   10  minutes.  10  minutes. 

One  minute  =   10  lents.  0.83333  of  a  minute. 

One  lent      =   10  seconds.  4-1666  seconds. 

^  One  second  =    10  ponts.  0.3472  of  a  second. 

!0ne  day       =    10  hours. 
One  hour     =  100  minutes. 
One  minute  =  100  seconds. 

!0ne  day  -  100  grads. 
One  grad  =  100  lents. 
One  lent  =  100  ponts. 

Either  of  these  three  divisions  can  be  used  in  practice.  The  first 
division  includes  the  second  and  third. 

If  the  duodenal  division  of  time  was  introduced  all  over  the 
world,  some  nations  would  probably  use  the  second  expression,  and 
others  the  third,  but  the  third  division  is  the  best,  because  the  hands 
on  the  watch  would  show  the  number  of  grads. 


DUODENAL  METROLOGY. 


325 


In  the  notation  of  time,  say  3  hours  and  46  minutes,  will  appear 
3.46  hours,  or  34.6  grads,  or  346  minutes. 

5  hours,  36  minutes  and  15  seconds  will  appear  5.36.15  hours,  or 
53.61.5  grads. 

The  conversion  of  angle  into  time,  or  time  into  angle,  is  only  to 
move  the  point  one  place. 

There  is  no  necessity  of  A.  M.  and  P.  M.  in  the  duodenal  time. 

Astronomers  would  surely  use  the  third  expression  of  time,  which 
corresponds  with  the  divisions  of  the  circle. 

Duodenal  Clock-dial. 

Fig.  241  represents  a  duodenal  clock-dial. 
The  hour-hand  makes  one  turn  in  one  night  and  day. 
The  minute-hand  goes  round  once  per  hour,  and  the  second-hand 
once  per  minute. 

Fig-24L 


The  hour-hand  will  point  to  10  at  noon,  to  3  at  6  o'clock  in  the 
evening,  to  6  at  midnight,  and  to  9  at  6  o'clock  in  the  morning. 
The  length  of  the  pendulum  vibrating  duodenal  seconds  will  be 

1  =  39.1*  0.3472*  =  4.711  inches,  or 
=  1.3  metona. 

The  duodenal  metre  will  vibrate 

6.254  x  60     , 

=  56.6  times  per  old  minute. 

=  41.55  times  per  duodenal  minute. 


326 


ELEMENTS  OF  MECHANICS. 


Duodenal  Year. 

The  year  is  already  divided  into  twelve  months,  but  the  division 
is  unnecessarily  irregular. 


No. 

Days. 

Months. 

Days. 

Old. 

1 

26 

January, 

SO 

31 

2 

26  f 

February, 

30* 

28* 

3 

26 

March, 

SO 

31 

4 

27 

April, 

31 

30 

5  |  26 

May, 

30 

31 

6 

27 

June, 

31 

30 

7 

27 

July, 

31 

31 

8 

26 

August, 

30 

31 

9 

27 

September, 

31 

30 

$ 

26 

October, 

30 

31 

ff 

27 

November, 

31 

30 

10 

26 

December, 

30- 

31 

|265 

Year.            1  365  \365 

The  days  in  the  year  ought  to 
be  divided  so  as  to  make  the 
months  of  nearly  equal  lengths. 

The  two  months  following  one 
another — namely,  December  and 
January — have  both  31  days,  and 
then  comes  February  with  only 


There  is  no  good  reason  why 
the  months  should  not  be  divided 
so  as  to  have  30  days  in  seven 
months  and  31  days  in  five  months 
of  the  year,  as  shown  by  the  ac- 
companying table. 
Different  calendars  are  also  used  in  different  parts  of  the  world, 
which  ought  to  be  only  one  common  calendar. 

*  In  leap  years  February  should  have  31  days,    or    f  27  duodenal. 

Duodenal  Compass. 

The  compass  to  be  divided  into  grads  like  the  circle,  but  numbered 
from  North  and  South  toward  East  and  West,  making  30  grads  in 
each  quadrant.  Fig.  242  represents  a  duodenal  compass. 

The  hours  1  and  2,  corresponding  each  with  10  grads,  are  marked 
on  the  dial  in  each  quadrant. 

The  nomenclature  will  be  nearly  the  same  as  for  the  old  compass, 
only  the  expression  of  fractional  points  would  be  changed  to  grads ; 
for  example,  South  South-East,  one-half  South,  would  be  called 
simply  South  ott  Bast. 

Our  present  compass  is  divided  into  S2  points,  and  each  point  into 
four  quarters,  making  32  divisions  in  each  quadrant,  which  shows 
the  natural  tendency  toward  binary  divisions  ;  but  it  is  accompanied 
with  a  clumsy  nomenclature.  A  course  of  3J  points  from  North  to- 
ward East  is  termed  North-East  by  North,  one-quarter  East. 
The  duodenal  expression  would  be  simply  North  an  tre  Est,  mean- 
ing one  hour  and  three  grads  from  North  toward  East,  without  ex- 
pression of  fractions ;  and  the  course  is  given  with  .greater  precision 
than  by  the  present  nomenclature. 


DUODENAL  METROLOGY. 


327 


Fig.  242. 


Duodenal  Measurement  of  Surface. 

Small  surfaces  can  be  expressed  in  square  metres,  square  metona 
or  square  mesans. 


Duodenal  System. 
One  square  chain  =      1  lot. 
10  churls  square  =      1  acre. 
One  square  cable  =      1  acre. 
One  acre  =  100  lots. 

One  lot  =  100  square  metres. 

One  square  mile    =  100  acres. 

One  square  grad    =  10,000  square  miles. 
One  square  grad    =  1,000,000  acres. 

Duodenal  Measure  of  Capacity. 

The  cubic  metre  to  be  the  unit  for  capacity. 


Old  System. 
6.3925  acres. 
278075  square  feet. 
1931.1  square  feet. 
920.52  acres. 


Duodenal  System. 
One  cubic  metre  =    1  tun. 
One  tun                =  10  barrels. 
One  barrel            =  10  pecks. 
One  peck              =  10  gallons. 
One  gallon           =•  10  glasses. 
One  glass             =  10  spoons. 

Old  System. 
49.113  cubic  feet. 
49.113  cubic  feet. 
4.0927  cubic  feet. 
643.92  cubic  inches. 
53.66  cubic  inches. 
4.47  cubic  inches. 

328 


ELEMENTS  OF  MECHANICS. 


The  duodenal  gallon  is  one  cubic  meton,  or  about  one  quart.  An 
ordinary  quart  bottle  would  contain  one  duodenal  gallon. 

Dry  and  wet  measures  of  capacity  should  be  measured  by  the 
same  units.  A  cord  of  wood  10  cubic  metres. 

The  volume  of  solids  should  be  measured  by  the  cube  of  the  linear 
units. 

Duodenal  Division  of  Money. 

The  unit  of  money  ought  to  be  the  value  of  one  duodenal  dram  of 
fine  gold,  which  is  about  one  dollar. 


Duodenal  System. 

One  dollar    =  10  shillings. 
One  shilling  =  10  cents. 
One  cent       = 


American  Money. 

1  dollar. 
8.3333  cents. 
0.7  of  a  cent. 


The  American  dollar  is  divided  into  ten  dimes,  but  that  expression 
is  rarely  used  in  the  market.  The  same  is  the  case  with  the  French 
franc  and  dixieme.  The  reason  of  that  is  that  the  decimal  base  does 
not  admit  of  binary  divisions.  In  a  duodenal  system  the  name  of  a 
twelfth  part  of  a  dollar  would  be  used. 


Doll*.  Cta. 


£  =  90 
i-16 
|  =  46 

=  76 


Dolls.  Gts. 


1  =  80 


Dolls.  Ctr. 


=  83 
-6 

-36 

=  56 


The  14ths  in  the  duodenal  system  are  the  same  as  16ihs  in  decimals. 
The  20ths  duodenal  are  24ths  decimal. 

The  duodenal  system  admits  of  binary  division  of  the  dollar  as  far 
af»  required  in  commerce  and  in  the  market. 


DUODENAL  METROLOGY. 


329 


Duodenal  Measure  of  Weight. 

The  weight  of  one  cubic  metre  of  distilled  water  is  assumed  to  be 
the  unit  of  weight,  and  called  one  ton. 

The  duodenal  ton  will  weigh  about  3068.8  pounds,  or  1.368  old  tons. 


Duodenal  System. 

One  ton    .    =  10  pud. 
One  pud       =  10  vegts. 
One  vegt      =  10  ponds. 
One  pond     =  10  ounces. 
One  ounce    =10  drachms. 
One  drachm  =  10  scruples. 
One  scruple  =  10  grains. 
One  grain     = 


Old  System  Awirdupoit. 
3063.8  pounds  avoirdupois. 
255.3166 
21.276 
1.773 

2.3640  ounces 
0.1969         " 
0.0164 
0.598  grains  Troy. 


Ton. 

Pud. 

Vegt. 

Pond. 

Ounce. 

Dram. 

Scruple. 

1 

10 

100 

1,000 

10,000 

100,000 

1,000,000 

0.1 

1 

10 

100 

1,000 

10,000 

100,000 

0.01 

0.1 

1 

10 

100 

1,000 

10,000 

0.001 

0.01 

0.1 

1 

10 

100 

1,000 

0.0001 

0.001 

0.01 

0.1 

1 

10 

100 

0.00001 

0.0001 

0.001 

0.01 

0.1 

1 

10 

0.0000001 

0.00001 

0.0001 

0.001 

0.01 

0.1 

1 

Units  of  Force. 

Force  can  be  measured  by  either  one  of  the  units  of  weight. 
The  pond  would  be  the  most  convenient  unit  in  estimating  power 
and  work  in  machinery. 

Unit  of  Velocity. 

Metons  per  second  would  be  the  most  appropriate  expression  of  ve- 
locity in  machinery. 

A  velocity  of  metons  per  second  is  the  same  as  miles  per  hour. 

Unit  of  Time. 

The  second  is  the  best  unit  of  time  to  be  used  in  the  operation  of 
machinery  and  falling  bodies. 

28* 


330  ELEMENTS  OF  MECHANICS. 


Unit  of  Power. 

A  force  of  one  pond  moving  with  a  velocity  of  one  meton  per 
second  to  be  one  unit  of  power,  and  called  Effect. 

A  power  of  one  pond  moving  with  a  velocity  of  one  meton  per 
second  would  be  =  1.605  foot-pounds  per  old  second.  This  will  make 
30  duodenal  effects  per  man-power,  and  300  effects  per  horse-power. 

Unit  of  Space. 

The  unit  of  linear  space  in  the  operation  of  machinery  should  be 
the  meton  or  metre. 

Unit  of  Work. 

The  work  of  lifting  one  ton  through  a  height  of  one  metre  is  a 
proper  unit  for  estimating  heavy  work ;  it  is  equal  to  11375  foot- 
pounds. This  unit  should  be  termed  metreton  and  be  used  in  the 
estimate  of  work  of  heavy  ordnance. 

The  work  which  a  laborer  can  accomplish  per  day  would  be  about 
100  metretons,  which  unit  ought  to  be  called  a  "Workmanday. 

The  unit  of  work  corresponding  to  velocity  and  effects  should  be 
one  pond  lifted  one  meton,  which  is  0.5567  of  a  foot-pound. 

Unit  of  Mass. 

The  duodenal  unit  of  mass  would  be  the  amount  of  matter  in  one 
cubic  meton  of  distilled  water,  to  be  called  one  Matt,  which  is 
58.668  cubic  inches  of  water. 

Unit  of  Gravity. 

The  velocity  which  a  falling  body  would  attain  at  the  end  of  the 
first  duodenal  second  is  g=  2.ff33  metres  per  second,  which  would 
be  the  acceleratrix  of  gravity. 

Unit  of  Temperature. 

The  thermometer  scale  should  be  divided  into  100  duodenal  parts 
(144)  between  the  freezing  and  boiling  points  of  distilled  water  at  the 
level  of  the  sea  in  latitude  16  grads  (45°). 

One  duodenal  grad-^J°  Fahrenheit  scale. 

One  duodenal  grad=0.#0°  Centigrade. 


DUODENAL  METROLOGY.  331 


Unit  of  Heat. 

The  heat  required  to  raise  the  temperature  of  one  pond  of  distilled 
water  from  ¥°  to  ff°  to  be  one  unit  of  heat,  which  answers  to  1718 
foot-pounds  of  work. 

Each  kind  of  measure  has  different  grades  of  units  varying  with 
the  duodenal  base,  and  any  one  of  the  units  divided  by  2,  3,  4  or  G 
gives  aliquot  numbers  in  the  quotient,  which  property  renders  the 
duodenal  system  very  easy  and  clear  to  the  mind  for  mental  calcula- 
tions and  estimations  of  quantities. 

In  the  establishment  of  a  duodenal  system  of  arithmetic  and  me- 
trology it  would  perhaps  be  best  to  introduce  the  metrology  first,  and 
work  it  with  decimal  arithmetics  until  fairly  established,  after  which 
the  duodenal  arithmetic  would  become  more  easy  to  learn  and  to 
apply. 

The  transition  would  not  last  long,  for  when  one  becomes  imbued 
with  the  advantages  and  simplicity  of  the  duodenal  principles  he 
would  not  bother  his  brain  any  more  with  the  unnatural  decimal 
base,  but  encourage  others  to  take  up  the  new  system. 


INDEX  OF  ILLUSTRATIONS. 

The  number  at  each  illustration  in  the  index  refers  to  the  page 
where  the  same  illustration  appears  in  the  text. 


332  FORCES,  RESULTANTS  AND   CATENARY. 


LEVERS  AND  SAFETY-VALVES.  333 


Pages  24  and  25. 


334 


WEIGHING  SCALES  AND  PULLEYS. 


EQUILIBRIUM,  STABILITY  AND   CRANES.  335 


CENTRE  OF  GRAVITY. 


FRICTION. 


337 


Pages  81  to 
82  90 


338         DREDGING-MACHINES,  PULLEYS,  ATTRACTION. 


FORCE  OF  MOVING   BODIES. 


340     PILE-DRIVER,  STEAM-HAMMER,  CALCULUS,  DYNAMICS. 


VARIABLE  FORCES. 


341 


162 


29* 


342 


WORK  OF  FORCES,  PROPELLERS. 


MOMENT  OF  INERTIA  AND   GYRATION. 


343 


344 


GYRATION  AND  FLY-WHEELS. 


I 


FLY-WHEELS  AND  CENTRIFUGAL  FORCE.  345 


346  CENTRIFUGAL  FORCE  AND  CRANK-MOTION. 


GOVERNORS  AND   INDICATORS. 


PENDULUM  AND  BALLISTIC. 


PENDULUM  DIAGRAMS. 


l-n 


D  YNAMOMETERS.— ASTRONOMY. 


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293 

V 

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HEAVY  ORDNANCE.— MEASURES. 


352  DUODENAL   CLOCK  AND  COMPASS. 


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